Topodex
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This "blog" project will hopefully develop into a complete set of notes on topology with an emphasis on low dimensional manifolds and Heegaard Floer homology. It exists primarily as a reference
for me when I need to remember where to find things.
Homological Algebra and Misc. Algebraic Topology
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Category Theory
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!!!
_Definition_ A concrete category is a category $\mathcal{C}$ together with a faithful functor $U : \mathcal{C} \rightarrow SET$.
A category for which such a functor exists is called concretizable.
Concrete categories can be interpreted as sets with additional structure. Examples include TOP, Ab, RMOD, etc. together with the forgetful functor.
Knot Theory
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A Primer on Geometric Topology
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First we cover general position. In smooth manifold theory, the intersection of two smooth submanifolds is guaranteed to be another smooth submanifold when the intersection is transverse. When two surfaces $\Sigma_1, \Sigma_2 \subset S^3$ meet transversely $\Sigma_1 \cap \Sigma_2$ is a smooth 1-dimensional submanifold of $S^3$ which can consist of finitely many loops and arcs. When
But when can we guarantee that a map is transverse? The Thom transversality theorem tells us that a smooth map $f : X \rightarrow Y$ can always be perturbed to be transverse to a given smooth submanifold $Z \subset Y$. We can immediately see some beautiful applications of this fact.
__Theorem__
_There are no non-orientable hypersurfaces in $\mathbb{R}^{n}$._
The following proof is due to Hans Samelson.
_Proof_
A non-orientable hypersurface is a codimension 1 submanifold of $\mathbb{R}^n$ which is closed (as a subset), without boundary, and non-orientable.
Suppose that such a hypersurface $H$ did exist. Then we may find a loop on $H$ such that the normal vector pushed smoothly along this loop returns back to itself with the opposite orientation.
By placing a point on the normal pushed off of $H$ and transporting it via the loop on $H$ and then connecting the two ends via the normal going through $H$ we obtain a new smooth closed path $\gamma \subset \mathbb{R}^n$ such that $\gamma$ meets $H$ at a single point. Since $\mathbb{R}^n$ is contractible, we can find a smoothly embedded disk $D \hookrightarrow \mathbb{R}^n$ whose boundary is $\gamma$.
By the transversality theorem, we can take this disk to be transverse to $H$ and hence $H \cap D$ will consist of finitely many closed loops in the interior of $D$ and a finite collection of arcs which have their end points on the boundary of $D$. However, the end points of these arcs must be distinct which tells us that $\partial D \cap H$ must consist of an even number of points, but we know that $\partial D \cap H$ consists of a single point, hence we have reached a contradiction. $\square$
As a corollary, we see that projective planes and Klein bottles may never be smoothly embedded into $\mathbb{R}^3$.
__Proposition__ The complement of a closed codimension one smooth submanifold of $\mathbb{R}^n$ is disconnected.
_Proof_ By the tubular neighborhood theorem, we know that this submanifold admits a tubular neighborhood that we identify as its normal bundle. We know that the normal bundle is locally trivial, then wlog let us choose two points $p$ and $q$ locally on opposite sides of $M$ (that is they live
in the same fiber of the tubular neighborhood). There is a unique segment $\alpha$ identified as the vectors in the fiber connecting them. Now assuming that the complement is connected, we can find a
smooth curve $\gamma \subset \mathbb{R}^n \backslash M$ connecting $p$ and $q$, then by the fact that $\mathbb{R}^n$ is simply connected, there exists a smooth disk $D$
with boundary $\partial D = \gamma \cup \alpha$ and by transversality $D$ can be taken to be transverse to $M$ but as in the argument of the previous theorem
we realize that the number of points in $\partial D \cap M$ has the wrong parity and hence we reach a contradiction. $\square$.
I leave it as an exercise to the reader to use this line of argument to show that the normal bundle of a smooth hypersurface in $\mathbb{R}^n$ is trivial (we've essentially already done this).
I should also mention that smoothness was necessary here as we needed a smoothly embedded disk to bound our simple closed curve in order to invoke transversality. I'm sure a similar argument can be
done in the locally flat setting but this is speculation at the moment.
__Jordan Curve Theorem__ Every simple closed curve in $\mathbb{R}^2$ seperates the plane into two pieces.
The following set of theorems are rather nontrivial. For proofs I recommend "Topology of 2 and 3 manifolds" by Moise.
__Theorem : __
_(Jordan-Schönflies) If $S^1 \hookrightarrow S^2$ is a smooth embedding, then $S^2 \backslash \iota(S^1) \cong \mathbb{D}^2 \sqcup \mathbb{D}^2$._
This tells us that any smoothly embedding circle in $S^2$ cuts out two hemispheres.
__Theorem : __
_(Schönflies) If $S^2 \hookrightarrow S^3$ is a smooth (PL) embedding, then $S^3 \backslash \iota(S^2) = \mathbb{B}^3 \sqcup \mathbb{B}^3$._
__Definition : __
_For any smoothly embedded 2-sphree $\Sigma \hookrightarrow S^3$ which meets $K$ in two points, then $K = A \sqcup B$,_
Next are the famous results of Papakyriakopoulos.
!!!
__Theorem : __
_(Sphere Theorem) If $Y$ is an oriented 3-manifold with $\pi_2(Y) \neq 0$, then there exists a properly embedded $S^2$ representing a non-zero class in $\pi_2(Y)$._
Let $K$ be a knot in $S^3$. If $\Sigma$ is a smoothly embedded 2-sphere which intersects $K$ transversely, then $\Sigma \cap K$ is an even collection of points. Suppose that $\Sigma \cap K$ consists of two points. The Schönflies theorem guarantees that $S^3$ is split into two sets homeomorphic to $B^3$. Let $\alpha$ and $\beta$ denote the arcs of $K$ on either side of $\Sigma$.
Joining a trivial arc going along $\Sigma$, we have now created two new knots $K_1$ and $K_2$. Notice that the knots $K_1$ and $K_2$ are independent of choice of arc.
!!!
__Definition__
_A knot is said to be prime if each smoothly embedded sphere $\Sigma$ which meets $K$ transversely in two points separates $K$ into components where one is unknotted._
A prime decomposition theorem exists for oriented knots. So that any knot can be uniquely decomposed into finitely many prime components.
One may also go the other direction by performing a connected sum of knots. Let $K_1$ and $K_2$ be two oriented knots in $S^3$. Isotope a strand from $K_1$ and one from $K_2$ so that one can find a smooth disk $D$ with oriented boundary consisting of four arcs, one of which on the strand of $K_1$ and one on the strand of $K_2$ (where the orientations align). Delete those two strands and then glue the other two strands from $\partial D$. The new knot $K_1 \# K_2$ inherits an orientation from $K_1$ and $K_2$. We can ask how properties of knots behave under connected sum.
!!!
__Theoreom__
_Let $K_1, K_2$ be knots in $S^3$, then $g(K_1 \# K_2 ) = g(K_1) + g(K_2)$._
_Proof_
Let $F_1$ and $F_2$ be genus minimizing surfaces for $K_1$ and $K_2$ respectively, then $F_1 \# F_2$ is a Seifert surface of $K_1 \# K_2$ with genus $g(F_1) + g(F_2)$ so we see that $g(K_1 \# K_2) \leq g(K_1) + g(K_2)$.
To see the other direction, let $F$ be a genus minimizing Seifert surface of $K_1 \# K_2$. Let $\Sigma$ be a 2-sphere separating $K_1$ and $K_2$, then we see by general position that $F \cap \Sigma = \beta \sqcup \left( \bigsqcup_{i= 1}^n \gamma_i\right)$ where $\gamma_i$ are simple closed curves. I claim that each innermost $\gamma_i$ bounds a disk $D \subset F$. If not, then we could cut along $\gamma_i$ and glue in two new disks which give a new Seifert surface for $K_1 \# K_2$ of genus lower than $F$, but since $F$ is genus minimizing, this is not possible and we see that $\gamma_i$ bounds a disk on $F$. We can then perturb $\Sigma$ to $\Sigma'$ so that $F \cap \Sigma'$ consists of one fewer $\gamma_i$. Repeating this process inductively, we see that there exists a 2-sphere $\Sigma'$ seperating $K_1$ and $K_2$ such that $F \cap \Sigma = \beta$ where $\beta$ is a simple arc between the two points in the set $K \cap \Sigma$. Cutting along $\beta$, we obtain two Seifert surfaces of $K_1$ and $K_2$ whose genuses sum to $g(K_1 \# K_2)$, hence $g(K_1 \# K_2) \geq g(K_1) + g(K_2)$ and we are done.
Algebraic Topology of the Knot Complement
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The exterior\footnote{Here we break slightly from the convention that 3-manifolds are denoted by $Y$. The notation $X_L$ is appropriate as we are studying the link \underline{EX}terior} is a connected compact orientable manifold with toroidal boundary, with the number of boundary components equal to the number of link components. It is clear that the homeomorphism type of $X_L$ is an invariant of the knot since any isotopy of $L$ induces a homeomorphism on $X_L$.
!!!
__**Proposition**__
*Let $L = \sqcup_{i=1}^n L_i$ be a $n$-component link in $S^3$,
$$
\widetilde{H}_k(X_L) = \begin{cases}
\mathbb{Z}^n \quad k = 1 \\
\mathbb{Z}^{n-1} \quad k = 2 \\
0 \quad \text{else}
\end{cases}
$$
moreover the generators of $H_1(X_L)$ are represented by meridians $\mu_i$ of each $L_i$, moreover if $L$ is an oriented link, then the orientations $\vec{\mu_i}$ are taken to be compatible with the ``right hand rule''.*
!!!
**Definition**
_A Seifert surface of $K$ is a connected compact orientable surface $F \subset S^3$ such that $\partial F = K$._
!!!
__Proposition__
_Let $X_K$ be the exterior of a knot $K \hookrightarrow S^3$, then $H_2(X_k, \partial X_k) \cong \mathbb{Z}$ and moreover, generators of $H_2(X_K, \partial X_K)$ correspond to Seifert surfaces of $K$ _
_Proof_
By Theorem \ref{linkhomology} it follows that
$$
H_n(X_K) = \begin{cases} \mathbb{Z} \quad n = 0,1 \\ 0 \quad \text{else} \end{cases}
$$
and that $\partial X_k$ is homeomorphic to a torus, thus
$$
H_n(\partial X_K) = \begin{cases} \mathbb{Z} \quad n = 0,2 \\ \mathbb{Z}\oplus\mathbb{Z} \quad n = 1 \\ 0 \quad \text{else} \end{cases}
$$
Now consider the following long exact sequence
$$
\require{AMScd}
\begin{CD}
\longrightarrow H_2(X_k) @>{}>> H_2(X_k, \partial X_k) @>>> H_1(\partial X_k) @>{\iota_*}>> H_1(X_k) @>{j_*}>> H_1(X_k, \partial X_k) \longrightarrow ;\\
@VVV @VVV \\
\end{CD}
$$
Notice that $\partial$ is injective. Since $\iota_*$ is the induced map by inclusion, we know that $\rm{im} \iota_*$ is the homology class generated by meridians of $K$, thus $\rm{im} \iota_* = [\mu] \cong \mathbb{Z}$, thus it follows that $H_2(X_k, \partial X_k) \cong \mathbb{Z}$ and moreover is generated by surfaces whose boundary corresponds to longitudes of $K$, in other words, generated by Seifert surfaces $F$ of $K$.
The following proposition follows via a similar argument.
__Proposition : __ _Let $X_L$ be the complement of an $n$-component link $L \hookrightarrow S^3$, then $H_2(X_L, \partial X_L) \cong \mathbb{Z}^n$ and generators of $H_2(X_L, \partial X_L)$ correspond to surfaces which cobound components of $L$_
So the homology type of $X_L$ is determined by the number of components of the link. Other obvious invariants would be the homotopy groups of $X_L$, in particular the fundamental group $\pi_1(X_L)$ which is called the knot (link) group. We discuss \hyperref[compalexpolynom]{methods to compute the link} group in Appendix \ref{compalexpolynom}, but for the moment what can we say about its properties? We know that $\pi^{ab}_1(X_L) = H_1(X_L) = \mathbb{Z}^n$. Additionally, $\pi_1(X_L)$ is finitely presented. Higher homotopy groups turn out to be trivial.
__Theorem : __
_Let $L$ be a non-split link in $S^3$, then $X_L = S^3 - L$ is a $K(\pi_1(X_L), 1)$ space. _
_Proof_
If $\pi_2$ is non-trival, then by the sphere theorem, there exists an element $[f: S^2 \rightarrow X_L] \in \pi_2(X_L)$ that admits an embedding $\Sigma$. Since this embedding misses $L$ we can also think of it as an embedding into $S^3$. By the Alexander-Schönflies theorem it follows that this embedding bounds two $B^3$'s: $S^3 = B^3_1 \cup_{\Sigma} B^3_2$. Since $L$ is non-split, we may assume WLOG $L \subset B^3_1$, but then $\Sigma = \partial B^3_2$ in $S^3 - L$, thus $\Sigma$ is nullhomotopic in $X_L$, and thus $\pi_2(X_L) = 0$. Since $\overline{S^3 - L} \simeq S^3 - L$ is open, it follows from Proposition 3.29 in Hatcher\cite{hatcher} that $H_3(X_L) = 0$. Since $H_i(X_L) = 0$ for all $i \geq 2$ it follows from the Hurewicz theorem that $\pi_i(X_L) = 0$ for all $i \geq 2$.
__Proposition : __
_Non-split link exteriors are Haken, and in particular knot exteriors are Haken_
_Proof_
Seifert surfaces are compact properly embedded surfaces. If $F$ is a genus minimizing Seifert surface, then $F$ is incompressible. It remains to show that the exteriors of non-split links are prime. This follows from the fact that non-split link exteriors are irreducible since any embedded sphere $S^2 \hookrightarrow S^3 - L$ is the boundary of a $B^3$ after applying the Alexander-Schönflies theorem. Since irreducible manifolds are prime, the result follows.
How strong is the homeomorphism type of $X_L$? If $L$ is a knot, then it is in fact a complete invariant! A famous theorem of Gordon and Luecke states that $X_{K_1} \cong X_{K_2}$ if and only if $K_1 \sim K_2$.
__Theorem__
_(Gordon-Luecke) Let $K_1$ and $K_2$ be knots in $S^3$, then $K_1 \sim K_2$ if and only if $X_{K_1} \cong X_{K_2}$_
This fails at the level of links.
__Proposition : __
_The 2-component link $L_n$ formed by aking $2n$-twisted knot with an unknot going through it. $L_n \neq L_m$ for $n \neq m$, however $X_{L_n} \cong X_{L_0}$ for all $n$._
_Proof_
Consider a solid unknotted torus $V$ where the twist link $T_{2n}$ runs parallel to its core. Notice that $W= S^3 - V$ is another solid torus where the unknotted component of $L_n$ forms its core. If we consider $W - L_n = \partial W \times [0,1) = \partial V \times [0,1)$, then applying a homeomorphism to $V$ which twists it by an $2n \pi$ rotation extends to a homeomorphism from $V - L_n$ to $V - L_n \cup W - L_n$. Notice that this homeomorphism changes $L_{n} $ to $L_{n + m}$. Thus the link complements are homeomorphic, however the Alexander polynomial distinguishes $T_{2n}$ for all $n$, hence they do not represent the same link up to ambient isotopy.
### $\pi_1(X_K)$: The knot group
First let us cover the classical Wirtinger presentation of the knot group.
Let us also recall that the
__Proposition : __
_ $\pi_1(X_K)$ is torsion free._
_Proof_
Suppose $a \in \pi_1(X_K)$ has order $n$, then there is a covering space $\widetilde{X}_K$ such that $\pi_1(\widetilde{X}_K) = \mathbb{Z}_n$, but then since $X_K$ is aspherical, $\widetilde{X}_K$ is also aspherical, hence $\widetilde{X}_K$ is a $K(\mathbb{Z}_n, 1)$ space, but such a space has non-zero homology groups $H_k$ for all $k$ odd, so we see that $\widetilde{X}_K$ is not a 3-manifold. $\square$
__Definition :__
_A knot $K$ is hyperbolic if the complement of $K$ in $S^3$ admits a hyperbolic structure (a Riemannian metric with constant negative sectional curvature)_
Jones Polynomial
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_Definition_ The Kauffman bracket associated to a link diagram $D$ is an integral Laurent polynomial $\left< D \right> \in \mathbb{Z}[A, A^{-1}]$ characterized by the following criterion:
![](https://timgbates.github.io/document/knottheoryimages/Kauffmanbrackerules.png width="450px" border="1")
The reader should ponder about whether or not this is well defined! We will prove this later, but for now
let us see how the bracket behaves under Reidemeister moves.
__Proposition__ The Kauffman bracket is unchanged by R2 and R3.
__Proposition__ The Kauffman bracket changes in the following way after R1.
In order to upgrade $\left< D \right>$ into an invariant, we need to choose an orientation for $K$.
__Proposition__ _The polynomial $(-A)^{-3w(D)} \left< D \right>$ is an invariant of the oriented link represented by $D$._
This invariant under a change of variables $A^{-2} \mapsto t^{1/2} $ is called the Jones Polynomial. One can check that the Jones polynomial is characterized by $V(O) = 1$ and
$$
t^{-1}V(L_+) - t V(L_-) + (t^{-1/2} - t^{1/2}) V(L_0)
$$
__Theorem__ _Let $K$ be an oriented knot, and $V_K(t)$ be its Jones polynomial, then $V_{\overline{K}} (t) = V_{K}(t^{-1})$. _
_Proof_
Handlebody Theory
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__Definition :__ _An $n$-dimensional $k$-handle $h$ is $\mathbb{D}^k \times \mathbb{D}^{n-k}$ which is attached to the boundary of an $n$-dimensional manifold $X$ along $\partial D^k \times D^{n-k}$ via $\varphi: \partial D^k \times D^{n-k} \rightarrow \partial X$._
+ $D^k \times 0$ is called the core
+ $0 \times D^{n-k}$ is called the cocore
+ $\partial D^k \times 0 = S^{k-1} \times 0$ is called the attaching sphere
+ $\partial D^k \times D^{n-k} = S^{k-1} \times D^{n-k}$ is called the attaching region.
+ $0 \times \partial D^{n-k}$ is called the belt sphere
+ $D^k \times \partial D^{n-k}$ is called the belt region.
__Lemma (Isotopy Lemma)__ _If $\varphi$ and $\varphi'$ are isotopic embeddings from $\partial D^k \times D^{n-k} \rightarrow \partial X$, then $X \cup_{\varphi} h \cong X \cup_{\varphi'} h$ _
__Lemma (Cancelation Lemma)__ _If $\varphi^k$ and $\varphi^{k+1}$ are a pair of k and (k+1)-handles such that the attaching sphere of $\varphi^{k+1}$ intersects the
belt sphere of $\varphi^k$ in a single point, then they can be cancelled._
__Theorem__ Every smooth compact n-manifold $M$ admits a handle decomposition, that is
__Proposition__ _If $M$ is connected, then we can eliminate all but one 0-handle from the handle decomposition. _
h-Cobordism Theorem
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__Definition__ _A cobordism between two oriented closed n-dimensional manifolds $M_0$ and $M_1$ is an (n+1)-dimensional compact oriented manifold $W$ such that
$\partial W = \overline{M_0} \sqcup M_1$._
The simplest cobordism that one can construct is the trivial cobordism from a manifold $M$ to itself, defined as $M \times [0,1]$. It is not hard to show that cobordisms form an equivalence relation
on oriented smooth closed manifolds of a given dimension.
__Definition__ _An h-cobordism between two manifolds $M_0$ and $M_1$ is a cobordism $W$ such that the inclusion maps $\iota: M_i \hookrightarrow W$ are homotopy equivalences._
Notice that if $M_i$ are simply connected, then the condition of $W$ being an h-cobordism is equivalent to $W$ being simply connected and $H_k(W, M) = 0$ for all $i$.
This follows from applying the Hurewicz theorem and CW approximation. An obvious h-cobordism from $M$ to itself is the space $M \times [0,1]$ from which we obviously can
conclude that both boundary components are diffeomorphic. What we will see is that admiting an h-cobordism in high dimensions is as good as the trivial cobordism.
__h-Cobordism Theorem__ _Let $M$ and $N$ be two compact simply connected oriented $m$-dimensional manifolds with $m \geq 5$ such that an h-cobordism $W$ exists,
then there exists a diffeomorphism $\varphi: W \rightarrow M \times [0,1]$ which may be taken to be identity on $M$, i.e. $\varphi|_{M} = 1_M$. _
The following presentation is taken from Milnor and showcases some powerful applications of the h-cobordism theorem.
__Theorem__ _Suppose $W^n$ is compact simply connected smooth n-dimensional manifold with $n \geq 6$ with simply connected boundary, then TFAE
1) $W$ diffeomorphic to $D^n$ 2) $W$ is homeomorphic to $D^n$ 3) $W$ is contractible 4) $W$ has the integral homology type of a point. _
_Proof_ It's obvious that 1 implies 2 implies 3 implies 4. It remains to show 4 implies 1. Take $D \subset W$ to be a smoothly embedded
disk in the interior of $W$, then $\partial (W - int(D)) = \partial D \sqcup N$, so $W - int(D)$ is a cobordism. Moreover $H_*(W - int(D), \partial D) \cong H_*(W, D) = 0$ by excision
hence $W - int(D_0)$ is an h-cobordism so by the h-cobordism theorem $W - int(D) \cong \partial D \times [0,1]$ and upon gluing in $D$ we see that $W$ is diffeomorphic to $D^n$ as required. $\square$
__Theorem (Poincare Conjecture)__ _Let $M$ be a homotopy n-sphere with $\dim M \geq 6$, then $M$ is homeomorphic to $S^n$. _1
_Proof_ Let $D$ be a smoothly embedded disk in $M$, then we
3-Manifold Topology
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Geometric-Facts
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!!!
__Theorem (Alexander)__ _Every closed oriented connected 3-manifold $Y$ is diffeomorphic to $Y'_{\phi} \cup_{Id|_{\partial Y' \times S^1}} \partial Y \times D^2$
where $Y'_{\phi}$ is the mapping torus of $\phi: Y'\rightarrow Y'$ a diffeomorphism of a connected compact orientable 3-manifold with boundary $\partial Y'$ such that
$\phi|_{\partial Y'} = Id$. _
An immediate corollary for this is the existence of foliations of 3-manifolds. Thurston and Winkelnkemper also use this to prove the existence of contact forms on closed orientable 3-manifolds.
__Definition__ Let $\Sigma \subset Y$ be an embedded closed surface. A compression disk for $\Sigma$ is an embedded disk $D \subset Y$ such that $\partial D \subset \Sigma$ and $\partial D$ does not bound a disk in $\Sigma$.
A non-sphere surface $\Sigma$ is said to be incompressible if it does not admit any compression disks.
A sphere $\Sigma \subset Y$ is called essential if it does not bound a 3-ball in $Y$.
__Theorem (Loop Theorem)__ _Let $Y$ be a 3-manifold with boundary $\partial Y$. If $(D^2, \partial D^2) \rightarrow (Y, \partial Y)$ is not null-homotopic, then
there exists an embedding $(D^2, \partial D^2 ) \hookrightarrow (Y,\partial Y)$ also not null-homotopic. _
An easy corollary is Dehn's lemma.
__Theorem__ (Dehn's Lemma)
Heegaard Decompositions
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Handlebody theory tells us that any smooth closed orientable 3-manifold $Y$ admits a handle decomposition.
Assuming this manifold is connected, there will be a unique 0-handle and a unique 3-handle. Since handles can be attached in ascending order we can begin with the 1-handles.
3-dimensional one handle is a copy of $\mathbb{D}^1 \times \mathbb{D}^2$ attached to the boundary of the 0-handle along $S^0 \times \mathbb{D}^2$. There is a unique framing for the attaching map that leaves the resulting space oriented.
After attaching the 1-handles, we are left with a space $H \cong \natural_g S^1 \times \mathbb{D}^2$. Next we need to see how the two handles are attached and since $\pi_1(O(1)) = 0$, it follows that we need only specify the isotopy class of the attaching sphere of this map.
These can be drawn in $\partial H_1 = \Sigma_g$. As there is a unique way to attach the 3-handle, it follows that a 3-manifold can be specified by a collection of $g$ embedded curves in $\Sigma_g$. Alternatively, turning the handlebody upsidedown, the 2-handles now becoming one handles, the 3 and 2 handles form a handlebody $H'$ that is homeomorphic to $\natural_g S^1 \times \mathbb{D}^2$. It remains to glue the two pieces together along their common boundary $\Sigma_g$, i.e. $Y = H \cup_{\varphi} H'$. We could specify an element in the mapping class group of $\Sigma_g$ or we can study where the belt spheres of the handles are mapped to in $\Sigma_g$. Consider the g-handle $H$ and remove a neighborhood of the belt spheres of each 1-handle. The space we are left with is homeomorphic to $\mathbb{D}^3$. We can then attach the neighborhoods of the belt spheres to the other handlebody $H'$ by specifying an isotopy class for each belt sphere in $\Sigma_g$ and then there is a unique way to glue in the 3-ball. Doing this for both handlebodies allows us to specify the 3-manifold.
These curves are called alpha and beta curves and they clearly satisfy:
+ The alpha curves are pairwise disjoint
+ Each alpha curve bounds a disk in H and each beta curve bounds a disk in H'
+ There are precisely $g$ alpha curves and $g$ beta curves
Dehn Surgery
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Closely related to the previous section is a procedure known as Dehn surgery.
__Lemma : __ _Let $Y$ be a closed oriented 3-manifold. Let $Y^0 = Y \backslash D^2 \times S^1$ for some embedded solid torus and $Y'$ the manifold obtained via $Y^0 \cup_{\varphi} D^2 \times S^1$ where $\varphi : S^1 \times S^1 \rightarrow S^1 \times S^1$, then $Y'$ is uniquely determined by where a meridian $\mu$ is mapped to via $\varphi$. _
Let $L$ be an oriented link in $S^3$. Along each component $L_i$ there is a unique isotopy class of curves in $\partial \nu(L_i)$ which has linking number 1 with $L_i$. Let $\mu_i$ denote this class. Next there is a unique longitude $\lambda_i$ which has algebraic intersection number $1$ with $\mu_i$. These are the canonical meridians and longitudes. If we remove an open neighborhood of $L_i$, and glue in a solid torus, we need only specify where the meridian of the solid torus is mapped to by the glueing map. As we only need to specify this up to isotopy, it suffices to describe this with $p \mu_i + q \lambda_i$. The reduced fraction $p_i/q_i$ is enough to describe the surgery on each component.
Plumbed 3-Manifolds
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Let us recall that there is a bijection between (oriented) disk bundles over $S^2$ and $\pi_1(SO(2)) \cong \mathbb{Z}$. We should then further recall that
each disk bundle has a distinct Euler number. A plumbed 4-manifold consists of a collection of disk bundles over $S^2$ and then
4-Manifold Topology
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Intersection Forms
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!!!
__Definition__ Let $X$ be a closed orientable 4-manifold. Define $Q_X : H^2(X;\mathbb{Z}) \times H^2(X ; \mathbb{Z})\rightarrow \mathbb{Z}$ to be the
symmetric bilinear where $Q_X(\alpha , \beta) = \left< (\alpha \cup \beta) , [M] \right>$. $Q_X$ is called the intersection form, and notice that
since $H_2(X ; \mathbb{Z}) \cong H^2(X ; \mathbb{Z})$ by Poincare duality, we can also define $Q_X : H_2(X ; \mathbb{Z}) \times H_2(X ; \mathbb{Z}) \rightarrow \mathbb{Z}$.
We need to justify the name _intersection form_ . First is the following useful lemma
!!!
__Theorem__ Let $X$ be a smooth closed orientable 4-manifold.
Let $\alpha \in H_2(X ; \mathbb{Z})$, then $\alpha = [\Sigma]$ for some embedded surface $\Sigma \subset X$.
We can then show that $Q_X(\alpha, \beta)$ records the algebraic signed intersection number of $\Sigma_{\alpha}$ and $\Sigma_{\beta}$.
__Proposition__ $Q_X$ is unimodular.
_Proof_ Without loss of generality let $H_2(X;\mathbb{Z})$ be torsion free.
We need to show that for any the map $\alpha \mapsto Q(\alpha^*, -)$ is an isomorphism from $H_2(X)$ to $H^2(X)$. Indeed this map can be written down
as $\alpha \mapsto \alpha^* \cup - $
!!!
The following are invariants of unimodular forms:
- The rank of $Q_X$ is an invariant which is equal to the second betti number $b_2(X)$.
- The signature of $Q_X$ is defined as the difference of positive and negative Eigenvalues over the real symmetric bilinear form $Q_X \otimes \mathbb{R}$.
The number of positive (negative) eigenvalues is written $b_2^+(X)$ ( $b_2^-(X)$ )
- The
!!!
__Lemma__ _Let $Q$ be a unimodular form on $\mathbb{Z}^n$, then if $v \in \mathbb{Z}^n$ is primitive i.e. $v = kv'$ implies $k = \pm 1$,
then there exists a dual element $w$ such that $Q(v,w) = 1$._
_Proof_ Let $A$ denote the matrix representation of $Q$ so that (in some basis) $Q(x,y) = x^T A y$. Then since $Q$ is unimodular
$A$ is invertible over $\mathbb{Z}$. Notice that $Av$ is primitive since $v = A^{-1}(Av) = A^{-1}(kw) = k A^{-1}w$ implies $ k = \pm 1$.
Now $Av = (\alpha_1, \dots , \alpha_n)$ is some primitive element in $\mathbb{Z}^n$ if and only if (\alpha_1, \dots , \alpha_n) = 1$.
Since $\mathbb{Z}$ is a Bezout domain the equation
$$
\alpha_1 x_1 + \alpha_2 x_2 + \dots + \alpha_n x_n = 1
$$
has an integral solution. This solution $w = (x_1, \dots , x_n)$ is the desired dual element. $\square$
The geometric interpretation is that for any surface $\Sigma$ representing a primitive class in $H_2(X)$, there exists a dual surface $\Sigma'$
with algebraic intersection number $1$.
__Definition__ Two symmetric bilinear unimodular forms on a free $\mathbb{Z}$-module $M$ are equivalent if and only if there exists an isomorphism $\Phi: M \rightarrow M$
such that $Q = \Phi^{-1}Q'$.
This is equivalent to the existence of a unimodular matrix $A$ such that $Q' = A Q A^T$.
Kirby Diagrams and Elementary Kirby Calculus
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
Vector Bundles
===============================================================================
!!!
__Definition__
_A vector bundle $\xi = (E, X, \pi)$ is a triple consisting of topological spaces $E$ and $X$ together with a continuous map $\pi: E \rightarrow X$ such that $\pi^{-1}(x)$ is a vector space and such that for all $x \in X$, there exists an open neighborhood $U \ni x$ and homeomorphisms $h: \pi^{-1}(U) \rightarrow U \times \mathbb{F}^n$
such that $h|_{\pi^{-1}(x')}$ is a linear isomorphism for all $x' \in U$._
$E$ is typically refered to as the total space, while $X$ is called the base space. When it is understood, often we refer to the vector bundle via its total space i.e. $E$ instead of $\xi$.
!!!
__Definition__ Two vector bundles $E$ and $E'$ over $B$ are said to be equivalent if there exists a homeomorphism $E \rightarrow E'$ which restricts to a linear isomorphism on each fiber.
!!!
__Definition__ _A section of a vector bundle $E \rightarrow B$ is a map $\sigma: B \rightarrow E$ such thata $\pi \circ \sigma = 1_B$. $\Gamma(E)$ refers to the sapce of
sections of $E$. _
If $E = B \times V$ is the trivial bundle, then a section is merely a continuous map from $B$ to $V$. Similarly, if $f : B \rightarrow V$ is
continuous, then we can construct a section $\sigma : B \rightarrow E$ defined by $f \mapsto (-, f(-))$. From this point of view a section is a generalization of the
space $C(B, V)$.
!!!
__Theorem__ If $Y \subset B$ is closed and $B$ is paracompact (admits partitions of unity) then for a vector bundle $E$ over $B$, the restriction map $\Gamma(E) \rightarrow \Gamma(E|_Y)$ is surjective.
The space $\Gamma(E)$ is also easily seen to be a vector space with operations $(\sigma + \tau) (x) = (x, \sigma(x) + \tau(x))$. It is then
easy to verify that the space of sections is contractible.
!!!
__Theorem__ _A rank n vector bundle $E$ admits n linearly independent sections if and only if $E$ is isomorphic to the trivial rank n bundle._
This gives us a second proof that the Mobius bundle is nontrivial, for any section must have positive and negative vectors and then the intermediate value theorem
implies that there must be a point where the section is zero.
Example: Recall from differential topology that the k-fold tensor product of the cotangent bundle is a smooth vector bundle over M,
and the sub-bundle of alternating k-tensors forms a vector bundle, a section of which is a k-differential form.
Construcing Vector Bundles
-------------------------------------------------------------------------------
!!!
__Definition__ Let $f: X \rightarrow Y$ be continuous and let $E \overset{\pi}{\rightarrow} Y$ be a vector bundle, then
define the pullback bundle $E_f \overset{\pi_f}{\rightarrow} X$ to be the subspace of $X \times E$ where $(x, v) \in E_f$
if $f(x) = \pi(v)$.
_Proof that $E_f$ is a well defined vector bundle_
Let us check that this is a vector bundle over $X$. Indeed $\pi_f^{-1}(x) = \{ (x, v) \mid \pi(v) = f(x)\}$ i.e. consists of the fiber over $f(x)$ in $E$,
hence this space naturally inherits a vector space structure over its fibers. It remains to check that this space admits a local trivialization.
Let $x \in X$ and let $U \subset Y$ be a neighborhood of $f(x)$ such that there exists a local trivialization $h: \pi^{-1}(U) \rightarrow U \times \mathbb{R}^n$.
Now let us notice that $V = f^{-1}(U)$ is an open neighborhood of $x$ and now let us define the map $\Psi: V \times \mathbb{R}^n \rightarrow \pi^{-1}(V)$
by $(v, c) \mapsto (v, h^{-1}(f(x), v))$. Clearly this map is continuous and bijective in the first factor and by the fact that $h$ is a local trivialization we see thata in the
second factor $\Psi$ is a linear isomorphism on each fiber and continuous, hence $\Psi$ defines a local trivialization on $V$ and hence we see that $E_f$ is a well defined vector bundle. $\square$
All of our favorite ways to build vector spaces also let us build vector bundles. Let $E$ and $F$ be vector bundles over the same base space $X$, then
- $E \oplus F$ is the vector bundle whose fibers are direct sums $E_x \oplus F_x$.
- $E \otimes F$ is the vector bundle whose fibers are tensor products $E_x \otimes F_x$.
- $Hom(E, F)$ is the vector bundle whose fibers are the vector spaces of linear maps from $E_x$ to $F_x$.
- $\Lambda^k(E)$ is the vector bundle whose fibers consist of alternating k-forms over $F_x$.
If $F$ is the trivial one dimensional bundle over $X$, then $Hom(E, \varepsilon^1)$ is called the dual bundle of $E$.
Notice that if $E$ can be equipped with a metric, then we can find an isomorphism from $E^*$ to $E$.
_Proof_ A finite dimensional vector space equipped with an orthonormal basis $\{e_1, e_2, \dots , e_n\}$ can be identified with
the dual space by the map $v = a_1 e_1 + a_2 e_2 + \dots + a_n e_3 \mapsto a_1 \epsilon^1 + \dots + a_n \epsilon^n$ where $\epsilon_i \in Hom(V, \mathbb{R})$ is
characterized by $\epsilon_i( v) = a_i$. Now let us call this isomorphism $h$. Now since we can equip $E$ with a metric, we can equip each fiber
with an orthonormal frame which varies continuously, then we can define the map $H : E \rightarrow E^*$ defined by $h$ on each fiber.
This map is continuous and an isomorphism on each fiber, thus we obtain a vector bundle isomorphism from $E$ to $E^*$. $\square$.
Notice that if the base space $B$ is paracompact, then we can find a partition of unity which allows us to construct a Euclidean metric on any vector bundle covering $B$.
In particular we see that if $B$ is a manifold, then any vector bundle is isomorphic to its dual.
!!!
__Theorem__ Let $E$ and $F$ be vector bundles over $X$, then there is a bundle isomorphism $E^* \otimes F \cong Hom(E, F)$.
_Proof_ Let $V$ and $W$ be finite dimensional vector spaces. Recall that there is a vector space isomorphism $\varphi: V^* \otimes W \rightarrow Hom(V,W)$ defined by
$v^* \otimes w (\cdot) \mapsto v^*(\cdot) w$. It remains to show that we can find a continuous map of vector bundles such that the restriction to
the fibers recovers this isomorphism.
!!!
__Theorem__ _Let $E$ be a vector bundle over $B \times [0,1]$ with $B$ compact, then the bundle $E|_0 = \{ ((b, 0), v) \in E\}$ and $E|_1 = \{ ((b,1), v) \in E\}$ are
isomorphic as vector bundles._
__Corollary__ _Let $f_0, f_1 : X \rightarrow Y$ be homotopic, and let $E \rightarrow Y$ be a vector bundle, then $f_0^*(E) \cong f_1^*(E)$._
_Proof_ Let $F : [0,1] \times X \rightarrow Y$ be a homotopy between $f_0$ and $f_1$. Then the bundle $F^*(E)$ restricted to $0 \times X$
and $1 \times X$ respectively are the bundles $f_0^*(E)$ and $f_1^*(E)$, so by the previous theorem they must be isomorphic. $\square$
__Corollary__ _If $B$ is contractible, then every vector bundle $E \rightarrow B$ is trivial._
__Proposition__ If $E \rightarrow Y$ is trivial, then $f^*(E)$ is trivial for any $f : X \rightarrow Y$.
_Proof_ Let $\sigma_1, \dots , \sigma_n$ be n linearly independent nowhere vanishing sections for $E$, then
define $\widetilde{\sigma}_i : X \rightarrow f^*(E)$ by $x \mapsto (x, \sigma_i(f(x)))$. Notice that $\widetilde{\sigma_i}$ are all nowhere vanishing
and linearly independent. $\square$
__Definition__ Let $E$ be a vector bundle over $X$, then the determinant line bundle is defined to be $\Lambda^n(E^*)$ where $n$ is the fiber dimension.
Thom Isomorphism Theorem and the Euler Class
-------------------------------------------------------------------------------
!!!
__Theorem__ Let $(E, B, \pi)$ be an oriented real rank n vector bundle, then there exists a class $u \in H^n(E, E_0 ; \mathbb{Z})$ called the
Thom class such that
the restriction of $u$ to any fiber maps $u$ to the orientation class $u|_F \in H(F, F_0;\mathbb{Z})$. Moreover, the map $\Phi : H^k(B)\rightarrow H^{k+n}(E,E_0)$
defined by $\alpha \mapsto \pi^*(\alpha) \smile u$ is an isomorphism.
Our first application will be the construction of the Gysin sequence of an oriented real vector bundle. First, there is a restriction map
on cohomology induced by the inclusion $(E, \emptyset) \rightarrow (E, E_0)$. There is also the natural isomorphism $\pi^* : H^*(B) \rightarrow H^*(E)$.
!!!
__Definition__ The Euler class $e(\xi)$ of an oriented real vector bundle is defined as the image of the restriction $u|_E$ of the Thom class
by $\pi^{-1*}$ .
!!!
__Theorem__ There exists a long exact sequence
$$
\begin{CD}
\longrightarrow H^j(B) @>{\smile e(\xi) }>> H^{j+n}(B) @>{\pi^*_0}>> H^{j+n}(E_0) @>{}>> H^{j+1}(B) \longrightarrow
\end{CD}
$$
called the Gysin sequence.
_Proof_ This sequences is really the long exact sequence on the pair $(E,E_0)$ in disguise. This sequences appears as
$$
\longrightarrow H^j(E,E_0) \longrightarrow H^j(E) \longrightarrow H^j(E_0) \longrightarrow H^{j+1}(E,E_0) \longrightarrow
$$
but recall the Thom isomorphism which yields $H^j(B) \cong H^{j+n}(E,E_0)$ by $\smile u$ with $u$ the Thom class. Replacing this part of the LES
yields $\longrightarrow H^{j-n}(B) \overset{\smile e(\xi)}{\longrightarrow} H^{j}(E)$. Since $E \simeq B$ via inclusion we see that $H^j(E)$ can be
replaced with $H^j(B)$ and that the last map can be replaced with $\pi^*_0$. $\square$
Stiefel-Whitney Classes
-------------------------------------------------------------------------------
!!!
__Definition__ Let $\xi = (E, B , \pi)$ be a real vector bundle and let $\Phi : H^k(B; \mathbb{Z}_2) \rightarrow H^{n+k}(E,E_0 ; \mathbb{Z}_2)$ be the Thom isomorphism,
and $u \in H^n(E,E_0;\mathbb{Z}_2)$ the Thom class, then the kth Stiefel-Whitney class is defined as
$$
w_k(\xi) = \Phi^{-1}(Sq^k(u))
$$
where $Sq^k$ is the kth Steenrod Square. $w(\xi) = 1 + w_1(\xi) + \dots + w_n(\xi)$.
Our goal now will be to show that this definition implies the following four axioms which characterize the Stiefel-Whitney classes:
1. $w_0(\xi) = 1$ and $w_i(\xi) = 0$ whenever $i$ is greater than the fiber dimension.
2. (naturality) Let $f : X \rightarrow Y$ be a continuous map which passes to a bundle map $E \rightarrow E'$, then $w_k(E) = f^*(w_k(E'))$
3. (Whitney Sum) Let $E$ and $F$ be two vector bundles over $B$, then
$$
w_k(E \oplus F) = \sum_{i+j = k} w_i(E) \smile w_j(F)
$$
4. $w(\gamma^1_1) \neq 1$.
First it is clear that axiom 1 follows as the ith Steenrod square for i larger than the fiber dimension is zero and $Sq^0(u) = \Phi^{-1}(1) = 1$.
Let $f : X \rightarrow Y$ be covered by a bundle map $E \rightarrow F$, then we can build a commutative diagram and the Thom isomorphism which will
show that $f^*(w_k(F)) = w_k(E)$.
Propery 3 follows from the sum formula of the Steenrod squares.
Property 4 follows by explicit computation. Notice that the total space of $\gamma_1^1$ deformation retracts to the Mobius bundle over $S^1$
and possesses boundary $S^1$ which itself is a deformation retract of $E_0$. If we take the boundary sum of this Mobius bundle with a disk, we obtain a copy of
$\mathbb{RP}^2$. Hence $H^*(E, E_0) \cong H^*(M, \partial M) \cong H^*(\mathbb{RP}^2, \mathbb{D}^2) \cong \widetilde{H}^*(\mathbb{RP}^2)$
is nonzero for $i = 1,2$. Thus the Thom class $u \in H^1(E,E_0)$ corresponds to the generator $a \in H^1(\mathbb{RP}^2)$ and so $u \smile u \neq 0$
and thus $w(\gamma_1^1) \neq 0$.
Lastly, the Stiefel Whitney class of a smooth manifold M written $w(M)$ is simply the Stiefel-Whitney class of its tangent bundle.
Notice that the total Stiefel Whitney class of a manifold $w(M) = 1 + w_1(M) + \dots + w_n(M)$ is a unit in the ring $H^{\prod}(M;\mathbb{Z}_2)$
of formal power series $1 + a_1 + a_2 + \dots$ with $a_i \in H^i(M;\mathbb{Z}_2)$. Let E and F be vector bundles over a space $B$ such that $E \oplus F = \varepsilon^N$ is trivial,
then $w(E) w(F) = 1$ so we can solve for $w(F)$ in terms of $w(E)$ by $w(F) = \overline{w(E)}$ where $\overline{w(E)}$ denotes the inverse element in the ring $H^{\prod}(M;\mathbb{Z}_2)$.
This yields an obstruction for a manifold embedding into $\mathbb{R}^N$.
!!!
__Definition__ (Normal Bundle) Let $M^k \hookrightarrow N^n$ be a smooth immersion, then $\nu(M)$ is the pullback of the subbundle of $TN^n|_{f(M)}$ consisting
of vectors $(f(x), v)$ with $v \cdot w = 0$ for all $w \in f_* TM$.
The normal bundle is well defined for smooth immersions by the injectivity of $f_*$ on $TM$.
!!!
__Proposition__ Let $M^k \overset{f}{\hookrightarrow} \mathbb{R}^n$ be a smooth embedding, then $TM \oplus \nu(M) = \varepsilon^n$ is trivial.
_Proof_ Notice that $f_* TM \subset T\mathbb{R}^n|_{f(M)}$ and that its complement is the normal bundle $\nu(M)$ so that $TM \oplus \nu(M) = f^*( T\mathbb{R}^n|_{f(M)})$. $\square$
Now if $\overline{w(M)} = 1 + \dots + a^k$, then $M^n \hookrightarrow \mathbb{R}^N$ implies that $N \geq n + k$.
As an example, recall that $w(\mathbb{RP}^4) = 1 + a + a^4$. It is not hard to determine that $1 + a + a^2 + a^3$ is the inverse so $\mathbb{RP}^4 \not\hookrightarrow \mathbb{R}^6$.
!!!
__Theorem__ The homomorphism $h : H^n(B;\mathbb{Z}) \rightarrow H^n(B;\mathbb{Z}_2)$ maps the Euler class of $\xi$ onto the top Stiefel-Whitney class $w_n(\xi)$.
_Proof_ Observe the following diagram,
$$
\begin{CD}
H^n(E,E_0;\mathbb{Z})@VVV @>>> H^n(E,E_0 ; \mathbb{Z}_2) @VVV\\
H^n(B ; \mathbb{Z} ) @>>> H^n(B;\mathbb{Z}_2)
\end{CD}
$$
and notice that the Thom isomorphism maps $ e(\xi) = u|_E$ to $u|_E \smile u = u \smile u = Sq^n(u)$ so that after reducing mod 2, the class $\Phi^{-1}(u \smile u \mod 2) = w_n(\xi)$.
!!!
__Theorem__ Let $E\rightarrow B$ be a real vector bundle over a paracompact space. $E$ is orientable if and only if $w_1(E) = 0$.
_Proof_ First, let us notice that the set of 2-sheeted covering spaces of B is in bijection with elements in $H^1(B; \mathbb{Z}_2)$.
This follows from the fact that 2-sheeted covering maps correspond via the galois corresondance to index 2 subgroups of $\pi_1(B)$ i.e.
to homomorphisms $[\pi_1(B), \mathbb{Z}_2] = [H_1(B), \mathbb{Z}_2]$ since $H_1(B)$ is the Abelianization of the fundamental group. Hence,
$Cov_2(B) \cong H^1(B ; \mathbb{Z}_2)$ and the claim is proved. Now let $P_{O}(E)$ be the principle $O(n)$ bundle (via assigning some Riemannian metric)
then we can quotient $P_O(E)$ be the action of $SO(n)$ to obtain a 2-sheeted covering space of B. Now let us assign to this two sheeted covering space the corresponding
class in $H^1(B;\mathbb{Z}_2)$. I claim that this is $w_1(E)$. One needs to only check that it is natural and that this yields the
correct class in the classifying space, and the rest follows from uniqueness of Stiefel-Whitney class. $\square$
__Corollary__ $\mathbb{RP}^n$ is orientable if and only if $n$ is odd.
_Proof_ Recall that $w(T\mathbb{RP}^n) = (1+a)^{n+1}$ so $w_1 = {n+1 \choose 1} = n+1 \pmod 2$ which is zero if and only if n is odd. $\square$
!!!
__Thereom__ Let $E \rightarrow B$ be an oriented real vector bundle, then $E$ admits a spin structure if and only if $w_2(E) = 0$.
Moreover, in the case that $w_2(E) = 0$, then there is a one to one correspondance between spin structures and $H^1(B; \mathbb{Z}_2)$.
__Corollary__ $\mathbb{CP}^2$ is not spin.
Complex Vector Bundles
-------------------------------------------------------------------------------
Complex vector bundles have some further peculiarities which make them stand out from the real case.
!!!
__Theorem (Splitting Principle)__ Let $E \rightarrow B$ be a complex vector bundle with B paracompact, then there exists a space $A$ and a continuous map $f : A \rightarrow B$ such
that $f^*(E) = L_1 \oplus L_2 \oplus \dots \oplus L_n$ splits along line bundles and furthermore that the map $f^* : H^*(B) \rightarrow H^*(A)$ is injective.
_Proof_ Let $\mathbb{P}(E) = \{ \ell \in F_b \subset E \}$ be the projectivization of E i.e. the fiber bundle over B where the fiber consists of the space of lines of the fiber of E.
It is easy to verify that this is a fiber bundle (local triviallity). There is also a surjection $H^*(\mathbb{CP}^{n-1}) \rightarrow H^*(F_b(\mathbb{P}(E)))$ on each fiber
which allows us to apply the Leray-Hirsh theorem.
A similar theorem (and proof) states the same for real vector bundles with $\mathbb{Z}_2$ coefficients in the cohomology ring.
Chern Pontrjagin Classes
-------------------------------------------------------------------------------
__Proposition__ The map $\otimes \mathbb{Z}_2$ takes the Chern classes $c_i(\xi)$ to the Stiefel Whitney classes.
_Proof_ Recall that this map sends the Euler class of a bundle $e(\xi)$ to the top Stiefel-Whitney class. Since $c_n(\xi) = e_{2n}(\xi)$, we
see already that this result is true for the top Chern and Stiefel-Whitney classes. For $i < n$, notice that in the Gysin sequence
$H^i(B) \rightarrow H^i(E_0)$ is an isomorphism, so that we see $e( \omega_0) \mapsto w_{n-1}(\omega_0) = w_{n-1}(E)$. Continuing inductively the result follows. $\square$.
Here is a cute application of Pontrjagin numbers.
__Proposition__ $\mathbb{CP}^2 \not\cong \overline{\mathbb{CP}^2}$.
_Proof_ Recall that the Pontrjagin numbers are unchanged by reversing orientations but the fundamental class
changes sign hence in order for there to exist an orientation reversing diffeomorphism, all Pontryagin numbers must
vanish, but we know that $p_1(\mathbb{CP}^n) = 3a^2 \neq 0$ and hence not all Pontryagin numbers vanish. $\square$.
Fiber Bundles
-------------------------------------------------------------------------------
!!!
__Theorem__ (Leray-Hirsch) Let $F \hookrightarrow E \rightarrow B$ be a fiber bundle such that $H^n(F; R)$ is a free R-module
and the restriction $\iota^* : H^n(E ; R) \rightarrow H^n(F ; R)$ is surjective, then the map
$$
H^*(B; R) \otimes H^*(F ; R) \rightarrow H^*(E ; R)
$$
defined by $\sum_{i,j} b_i \otimes \iota^*(c_j) \mapsto \sum_{i,j} \pi^*(b_i) \otimes c_j$ is an R-module isomorphism.
This makes computing the cohomology ring of some spaces straightforward. For example, there is a fibration $S^1 \hookrightarrow S^{\infty}/\mathbb{Z}_p \rightarrow \mathbb{CP}^{\infty}$
where of course $S^{\infty}/\mathbb{Z}_p = K(Z_p, 1)$. We can then use Leray-Hirsch to say the additive group structure of $H^*(K(Z_p, 1), \mathbb{Z}_p) = \Lambda(\alpha) \otimes \mathbb{Z}_p[\beta]$
with $|\alpha| = 1$ and $|\beta| = 2$. Of course one has to verify that the conditions for Leray-Hirsch are met.
Classifying Spaces
-------------------------------------------------------------------------------
!!!
__Theorem__ Let $\gamma^n$ be the tautological n-frame bundle over $Gr_n(\mathbb{R}^{\infty})$, then
there is a bijection $[X, Gr_n(\mathbb{R}^n)] \cong \textrm{Vect}^n(X)$ defined by $f \mapsto f^*(\gamma^n)$.
Vector bundles are thus classified by homotopy classes of
maps to the space $Gr_n(\mathbb{R}^{\infty})$ we we alternatively write as $BO(n)$ and call the classifying space for real vector bundles. This serves as a model theorem for what we will study now in more generality.
!!!
__Definition__ Let $G$ be a topological group. A principle G bundle is a fiber bundle $(P, X, \pi)$ equipped with a right G-action
on each fiber that is free and transitive.
These spaces are in some ways quite rigid. Any morphism of principle $G$ bundles (over the same base space) is an isomorphism of $G$-bundles. This tells us for example that if $P$ admits a section then it is trivial.
Indeed let $\sigma: B \rightarrow P$ be a section and then define $\psi: B \times G \rightarrow P$ by $(b, g) \mapsto (b, g\sigma(b))$. This map is clearly continuous and a bundle map and so an isomorphism.
!!!
__Definition__ Let $G$ be a topological group, and let $P \rightarrow X$ be a principle G bundle, then a classifying space BG is a
topological space equipped with a principle-G bundle $EG \rightarrow BG$ such that for any (paracompact) space X, there is a bijection
between principle G-bundles over X and homotopy classes of maps $[X, BG]$ where the bijection assigns an element $f \in [X, BG]$ to $f^*EG$.
!!!
__Proposition__ Let $BG$ and $BG'$ be two classifying spaces for G, then $BG \simeq BG'$.
_Proof_ This follows by the fact that there exists maps $f : BG \rightarrow BG'$ and $f' : BG' \rightarrow BG$ such that
$f^*EG' = EG$ and $f'^*EG = EG'$ implies that $f \circ f'$ is homotopic to identity. $\square$
Heegaard-Floer homology
===============================================================================
Heegaard-Floer homology is a collection of homology theories which are invariants of closed orientable 3-manifolds and of knots and links.
The construction of $HF$ is analytic and is the Lagrangian intersection Floer homology of the symmetric product of a Heegaard surface with
Lagrangians given by looking at the unordered tuples of points in the alpha curves and beta curves. In practice, Heegaard Floer possesses
a combinatorial structure which makes it more computationally friendly than some of its sister theories (such as Seiberg-Witten Floer and ECH).
4-dimensional cobordisms between three manifolds $Y_0$ and $Y_1$ induce maps $F_W : HF(Y_0) \rightarrow HF(Y_1)$ which topologists have expoited to
yield invariants of four manifolds.
$Sym^g(\Sigma)$
-------------------------------------------------------------------------------
!!! __Proposition__ $\textrm{Sym}^n(\mathbb{C}) = \mathbb{C}^n$.
_Proof_ Let $(a_1, a_2, \dots , a_n)$ be a tuple of n points in $\mathbb{C}$, then I can associate to this tuple a polynomial
$(z- a_1)(z-a_2)\dots (z-a_n) = z^n + c_1 z^{n-1} + c_2 z^{n-2} +\dots + c_{n-1} z + c_n$, where the $c_i$ are uniquely determined by the
collection $a_1, \dots ,a_n$ up to rearrangment. We see then that assigning an unordered n-tuple $(a_1, \dots ,a_n)$ to the tuple $(c_1, \dots , c_n)$ is
a well defined function from $\textrm{Sym}^n(\mathbb{C})$ to $\mathbb{C}^n$. I claim that this is a homeomorphism. Injectivity is obvious and surjectivity is a consequence of the
fundamental theorem of algebra. A little work shows that this map as well as its inverse are continuous. $\square$
__Theorem__ _Let $\Sigma$ be a closed oriented genus-g surface, then $\textrm{Sym}^g(\Sigma)$ is a 2g dimensional manifold. _
_Proof_ Since any surface is locally $\mathbb{C}$, we can use the previous proposition to construct local homeomorphisms from open subsets of $\textrm{Sym}^g(\Sigma)$ to open subsets of $\mathbb{C}^n$ and
then verifying that the transition maps are continuous. $\square$
Definition of Heegaard-Floer homology
-------------------------------------------------------------------------------
Let $Y$ be a closed oriented 3-manifold. We can associate a chain complex to a Heegaard decomposition of $Y$ in the following way. Let $(\Sigma, \alpha, \beta, z)$ be a tuple consisting of a closed oriented genus g surface, a collection of g disjoint closed curves $\alpha_i$ representing the belt spheres of one of the handlebodies whose boundary is $\Sigma$,
a collection of $g$ disjoint closed curves $\beta_i$ representing the belt spheres of the other handlebody in the Heegaard decomposition, and a point $z \in \Sigma - \mathbf{\alpha} - \mathbf{\beta}$ called the basepoint.
I claim that the space $\textrm{Sym}^g(\Sigma)$, the g-fold cartesian product of $\Sigma$ quotiented by the $S_g$ action interchanging the coordinates, is a manifold of dimension $2g$.
Inside this manifold there are half dimensional submanifolds $\mathbb{T}_{\alpha}$ and $\mathbb{T}_{\beta}$ which are defined to be the image of $(\alpha_1, \dots , \alpha_g)$ and $(\beta_1, \dots , \beta_g)$ after quotienting by $S_g$.
They represent Lagrangians inside of $\textrm{Sym}^g(\Sigma)$ (no I will not explain where the symplectic structure comes from because frankly I don't understand it). Note also that $z \times \Sigma \times \dots \times \Sigma$ defines a codimension 2 submanifold of $\textrm{Sym}^g(\Sigma)$. This means though that we can perform Lagrangian intersection Floer homology on the triple $(\textrm{Sym}^g(\Sigma), \mathbb{T}_{\alpha}, \mathbb{T}_{\beta})$. The chain complex will be the free abelian group generated by the points $x \in \mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}$. In Lagrangian intersection Floer homology we would now count the pseudo-holomorphic disks between a pair o points with Maslov index 1. Let $\phi(x,y)$ denote the homotopy classes of maps from the disk $\mathbb{D}$ into $\textrm{Sym}^g(\Sigma)$ such that $\phi(i) = y$ and $\phi(-i) = x$. Let $\mathcal{M}(\phi)$ denote the space of pseudo-holomorphic disks representing $\phi \in \pi_2(x,y)$. Since there is a 1-parameter family of holomorphic automorphisms of $\mathbb{D}$ which preserve $i$ and $-i$, the space $\overline{\mathcal{M}(\phi)}$ denotes $\mathcal{M}(\phi)$ quotiented by the action of this one-parameter family. The Gromov compactness theorem then guarantees that this space is finite. We can then define the differential $\widehat{\partial}$ to be
$$
\widehat{\partial}(x) = \sum_{ y \in \mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}} \sum_{\{ \phi \in \pi_2(x,y) \mid n_z(\phi) = 0, \mu(\phi) = 1 \}} \# ( \overline{\mathcal{M}(\phi)}) \cdot y
$$
where $n_z(\phi)$ is the algebraic intersection number of a generic representative of $\phi$ with the codimension 2 submanifold $z \times \textrm{Sym}^{g-1}(\Sigma)$. Taking homology of this chain complex is the definition of $\widehat{HF}(Y)$ and is invariant of the choices made along the way.
The next variant we discuss is the minus version of Heegaard Floer. We define $CF^-(\mathcal{H})$ to be the free $F[U]$ module generated by $\mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}$ with the differential defined to be
$$
\partial^- (x) = \sum_{y \in \mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}} \sum_{\phi \in \pi_2(x,y), \mu(\phi) = 1} \#(\overline{\mathcal{M}(\phi)}) U^{n_z(\phi)} \cdot y
$$
$U$ acts on $HF^-(Y)$ by dropping the homological grading by $2$. The attentive reader should be concerned that this map is well defined since $n_z(\phi)$ counts signed intersection of generic $\phi$ with $V_z$, but as both of these are complex submanifolds in $\textrm{Sym}^g(\Sigma)$ their intersection is guaranteed to be non-negative.
$CF^{\infty}$ is the chain complex freely generated over $\mathbb{F}[U,U^{-1}]$ with the same differential $\partial^-$. We can then define $CF^{+} = CF^{\infty}/CF^{-}$. We have a short exact sequence
$$
\require{AMScd}
\begin{CD}
0 @>>> CF^- @>>> CF^{\infty} @>>> CF^+ @>>> 0
\end{CD}
$$
This induces a long exact sequence on homology
$$
\require{AMScd}
\begin{CD}
HF^{-}(Y) @>>> HF^{\infty}(Y) @>>> HF^+(Y) @>>> HF^-(Y)
\end{CD}
$$
__Theorem__
_Let $Y$ be a rational homology sphere, then for all $\mathfrak{s} \in \textrm{Spin}^c(Y)$ _
$$
HFK^-(Y, \mathfrak{s}) = \mathbb{F}[U]_{(d)} \oplus \bigoplus_i \mathbb{F}[U]/(U^{n_i})
$$
### Gradings
Very crucial to the theory of Heegaard Floer homology (and Knot Floer homology) is the fact that the chain complex is bigraded.
First we define a relative grading $M$ on $\mathbb{T}_{\alpha}\cap\mathbb{T}_{\beta}$ called the Maslov grading as follows; let $x, y \in \mathbb{T}_{\alpha}\cap \mathbb{T}_{\beta}$, then $M(y) - M(x) = \mu(\phi) - n_z(\phi)$ with $\phi \in \pi_2(x,y)$.
Some work has to be done to show that this is well defined. One immediately notices that $\partial$ lowers the Maslov grading by one.
We also obtain a second relative grading on $\mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}$ induced by the spin-c structures on $Y$.
The differential $\partial$ respects this decomposition, so we can write
$$
\widehat{HF}(Y) = \bigoplus_{\mathfrak{s} \in \textrm{Spin}^c(Y)} \widehat{HF}(Y, \mathfrak{s})
$$
Notice that neither of these gradings are absolute gradings, only relative.
#### Spin-C Gradings
Let's define a map $s_z : \mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta} \rightarrow \textrm{Spin}^c(Y)$. Let $x \in \mathbb{T}_{\alpha}\cap \mathbb{T}_{\beta}$, then $x := (x_1, x_2, \dots , x_g)/S_g \in \textrm{Sym}^g(\Sigma)$. Let us choose a Morse function $f: Y \rightarrow \mathbb{R}_{+}$ which is compatible with the Heegaard decomposition.
The points $x_1, \dots , x_g \in \Sigma$ sit along $g$ flow lines of $\nabla f$ connecting pairs of index 1 and 2 critical points.
$z$ sits along a flow line $f_i$ connecting the unique index 0 and unique index 3 critical points. Taking a tubular neighborhood of the $g+1$ flow lines gives us a subset of $Y$ diffeomorphic to the disjoint union of $g+1$ balls, and moreover $\nabla g$ is a non-vanishing vector field on $Y - \cup_{1}^{g+1} \nu(f_i)$ and hence determines a $\textrm{Spin}^c$ structure on $Y$ (by the Turaev reformulation of Spin-c structures on 3-manifolds). This defines our map $s_z$.
__Proposition__ _If $s_z(x) \neq s_z(y)$, then $\pi_2(x,y) = \emptyset$. _
__Corollary__ _ $\partial$ respects the relative spin-c grading: $s_z(x) = s_z(\partial x)$ _
Let's now skip a couple of steps and talk about knot Floer homology.
__Definition__
_A doubly-pointed Heegaard diagram $(\Sigma, \alpha, \beta, w, z)$ is a Heegaard diagram with two points $w$ and $z$ in the complement of the $\alpha$ and $\beta$ curves. _
To a knot $K \subset Y$, we can associate a doubly pointed Heeegaard diagram by a curve in $\Sigma$ going from $w$ to $z$ which is disjoint from the alpha curves which we push into the alpha handlebody and a curve from $z$ to $w$ disjoint from the beta curves which we push into the beta handlebody. The union of these curves defines the knot $K$.
Given a knot $K \subset S^3$, there is a simple algorithm for constructing a doubly-pointed Heegaard diagram compatible with that knot. We will also
do this in such a way so that the generators $x \in \mathbb{T}_{\alpha}\cap \mathbb{T}_{\beta}$ are in one to one correspondance with the Kauffmann states
defined by Kauffmann in his `Formal Knot Theory' book. There are two gradings that we can put on the Kauffmann states which will correspond to the
Maslov and Alexander gradings. Ozsvath and Szabo exploited this to give a characterization of $\widehat{HFK}$ of alternating knots.
### Gradings
### Alternating knots
This should hopefully summarize the paper by OS on the knot floer homology of alternating knots.
__Theorem (Ozsvàth Szabò__ _Let $K$ be an alternating knot in $S^3$ and $a_s$ denote the $s$-coefficient in the symmetrized Alexander polynomial, then_
$$
\widehat{HFK}(K,s) = \mathbb{Z}^{|a_s|}
$$
_and is supported in Maslov grading $s + \sigma(K)/2$ where $\sigma(K)$ is the signature of the knot. _
This theorem is used to prove the next result.
__Theorem__ _Let $K$ be an alternating knot oriented such that $\sigma(K) \leq 0$. Then for all $s > 0$, _
$$
HF^+(K, s) \cong \mathbb{Z}^{b_s} \oplus \left( \mathbb{Z}[u]/U^{\delta(\sigma, s)}\right)
$$
What they do to prove the first theorem is construct a doubly pointed Heegaard diagram from a marked diagram of $K$ such that the Kauffman states are put in one to one correspondance with the generators of $\widehat{CFK}$. They then utilize functions $S : \mathscr{S} \rightarrow \mathbb{Z}$ and $M : \mathscr{S} \rightarrow \mathbb{Z}$ which are defined in terms of local contributions of the states then define gradings on elements in $\mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}$ which they prove correspond to the Maslov and Alexander gradings of the Knot Floer chain complex. Moreover, when the knot is alternating they obtain a formula
$$
2( M(x) - S(x)) = \sigma(K)
$$
where $\sigma(K)$ is the signature of the knot $K$. We see then that every generator supported in Alexander grading $s$ must be supported in Maslov grading $s + \sigma(K)/2$ and thus the homology of this complex in this grading is $\widehat{HFK}(K,s) = \mathbb{Z}^{b}$, where $b \in \mathbb{Z}_{\geq 0 }$. To determine $b$ they use the fact that
$$
\Delta_K(t) = \sum_{ x \in \mathscr{S}} (-1)^{M(x)} t^{S(x)} = \sum_{x \in \mathscr{S}} (-1)^{S(x) + \sigma(K)/2} t^{S(x)}
$$
where $\Delta_K(t)$ denotes the symmetrized Alexander polynomial (a theorem of Kauffman),
and so the rank of $\widehat{HFK}(K,s)$ is equal to $|a_s|$ where $a_s$ is the s coefficient in the symmetrized Alexander polynomial.
What is left for us to do is:
1. Construct this doubly pointed Heegaard diagram $\mathcal{H}$ and demonstrate that
- Kauffman states are in bijection with $\mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}$
- $\mathcal{F}(x) = S(x)$
- $\textrm{gr}(x) = M(x)$
2. Demonstrate that if $K$ is alternating then $2(M(x) - S(x)) = \sigma(K)$.
__Background__
Let $D$ be a marked diagram, and let $\mathcal{R}$ be the regions of the diagram excluding the two regions adjacent to the marked edge.
__Definition__ _A Kauffman state for a marked diagram $D$ is a one to one assignment of vertices to regions in $\mathcal{R}$. _
__First Step of Proof__
Let $D$ be a marked diagram for $K$ where the marked edge is adjacent to the region containing $\infty$. We can construct a Heegaard diagram by taking the surface which projects to a neighborhood of the diagram. This will be a genus $n+1$ surface. The $\alpha$ curves are those in this surface which trace out the boundaries of the regions (except the region containing infinity). Call $\alpha_{n+1}$ the curve which borders the region next to the marked edge. Let $\beta_1, \dots , \beta_n$ denote the curves which we locally get by
and $\beta_{n+1}$ is chosen to be a meridian around the marked edge, and place the $w$ and $z$ basepoints on either side of $\beta_{n+1}$. Notice that Kauffman states correspond to an assignment of a region (not including the adjacent regions to the marked edge) to a crossing, and that a generator in $\mathbb{T}_{\alpha} \cap \mathbb{T}_{\beta}$ corresponds to an assignment of an $\alpha$ curve to a vertex and the $\alpha_{n+1}$ curve to $\beta_{n+1}$ i.e. a Kauffman state.
$\mathcal{F}(x) = S(x)$
To show this we first say that two states are related by a transposition.
__Second Step__
Recall the theorem of Gordon and Litherland that if $K$ is alternating then
$$
\sigma(K) = \#\textrm{Black Regions} - \#\textrm{Positive Crossings} - 1
$$
We can see from the definitions of $m(x,v)$ and $s(x,v)$ that
$$
m(x,v) - s(x,v) + \frac{\epsilon(v) + 1}{4} = \begin{cases}
1/2 \quad \textrm{Over-under crossing} \\
0 \quad \textrm{Under-over crossing}
\end{cases}
$$
At this point I should point out that the OS paper has a TYPO. They have a 2 where I have a 4 in the expression above.
Now notice that
$$
\sum_{v} \frac{\epsilon(v) + 1}{2} = \frac{1}{2}[\#p - \# n + \#v] = \frac{1}{2} (2 \# p) = \#p
$$
and hence
$M(x) - S(x) + \#p = 1/2(\#\textrm{Black regions} - 1)$
we then see that
$$
2(M(x) - S(x)) = \#(\textrm{Black regions}) - 1 - \#\textrm{Positive crossings} = \sigma(K)
$$
Georgia Topology Summer School Notes
===============================================================================
I am fortunate enough to be attending the Georgia topology summer school! The emphasis this year is on surfaces in 4-manifolds.
Diagrams for Knotted Surfaces (taught by Mark Hughes)
-------------------------------------------------------------------------------
Since a large portion of this course was about representing knots with diagrams, I will instead link the slides that Mark provided for
this minicourse and for the remainder of this section I will give brief summaries and upload some examples that I come up with based on the lectures.
### Broken Surface Diagrams
### Banded unlink diagrams in 4-manifolds.
For this talk, $X$ will be a closed oriented smooth 4-manifold. Let $h : X^4 \rightarrow \mathbb{R}$ be a self-indexing Morse function.
Assume that $X$ has 1 0-handle. The way we're going to think about our handle decomposition is via the gradient vector field of our Morse functions.
Given $(L,B) \subseteq S^3 \backslash \nu(K)$, we can think of $(L,B)$ as living in either $M_{1/2}, M_{3/2}, M_{5/2}$.
https://arxiv.org/pdf/1804.09169 looks like a good reference for this.
### Braided Surfaces and Surface Braids.
Consequences of Disk Embedding Theorem (taught by Arunima Ray)
-------------------------------------------------------------------------------
The primary result that we will learn in this minicourse is the following theorem due to Lee and Wilczynski:
__Theorem__ _Let $X^4$ be a closed simply connected manifold, then any primitive class $\alpha \in H_2(X; \mathbb{Z})$ is represented by a locally lat torus._
We will also learn the following theorems due to Freedman and Quinn.
__Theorem__ _ 1) Any knot with trivial Alexander polynomial is topologically slice, 2) Any 2-knot $S^2 \hookrightarrow S^4$ with $\pi_1(S^4\backslash K) \cong \mathbb{Z}$ is topologically unknotted_
__Definition__ _An embedding $f: \Sigma \hookrightarrow M$ is locally flat if for every $x \in \Sigma$ there exists a neighborhood $U$ such that $(U, U \cap f(\Sigma)) \cong (\mathbb{R}^4, \mathbb{R}^2)$ _
__Definition__ _A map $f: \Sigma^2 \rightarrow M^4$ is called a generic immersion if it is a locally flat embedding except for isolated double point singularities i.e. there
exists homeomorphism $(U, U \cap f(\Sigma)) \cong (\mathbb{R}^4, \mathbb{R}^2_{xy} \cup \mathbb{R}^2_{zt})$ _
Next we need the following technical results due to Quinn.
__Theorem__ _(Immersion Lemma) Any map $\Sigma^2 \rightarrow M^4$ is homotopic to a generic immersion_
__Theorem__ _(Quinn) Any locally flat submanifold has a linear normal bundle.
and
If $\Sigma_1$ and $\Sigma_2$ are locally flat submanifolds of $M$ then there exists ambient isotopy of $M$ taking $\Sigma_1$ and $\Sigma_2$ to intersect transversely._
We can visualize double points using motion pictures i.e. sequences of copies of embeddings in $\mathbb{R}^3$ which we imagine as parameterizing $\mathbb{R}^3 \times \mathbb{R}$.
The Cliffford torus is the product of unit circles in $\mathbb{R}^2 \times \mathbb{R}^2$.
The last important result we need is the disk embedding theorem.
__Theorem__ [Freedman and Quinn] _Let $M^4$ be a topological manifold with trivial fundamental group and let $f : \mathbb{D}^2 \rightarrow M$ be a generic immersion with
$\partial \mathbb{D}^2 \hookrightarrow \partial M$ and let $g : S^2 \rightarrow M$ be a generic immersion such that $g$ has trivial normal bundle.
Let $g$ have trivial self intersection and $f, g$ be algebraically dual i.e. $f \cdot g = 1$. Then there exists $\overline{f} \simeq f$ rel $\partial f$
and $\overline{g} \simeq g$ rel boundary such that $\overline{f}$ is locally flat embedding and $\overline{f}$ and $\overline{g}$ are geometrically dual.
Having the `friend' $g$ is necessary in order for this theorem to hold. There are also various generalizations of the disk embedding theorem.
Now we start the proof of the main theorem.
### Step 0
Since $\pi_1(M) = 1$, the Hurewicz theorem tells us that $\pi_2(M) \cong H_2(M)$ so every primitive class $\alpha \in H_2(M)$
can be represented by an immersed sphere. We then use the immersion lemma to homotope this to a generic immersion. Since $\alpha$ is
primitive we can use the Poincare duality isomorphism to produce $\beta = [g : S^2 \rightarrow M]$, a generic immersion, which is algebraically
dual to $\alpha$. The geometric Casson lemma tells us that up to homotopy we can take $f$ and $g$ to be geometrically dual.
### Step 1
Link homologies and knotted surfaces (taught by Kyle Hayden)
-------------------------------------------------------------------------------
taught by Andras Stipsicz
-------------------------------------------------------------------------------
Let $X$ be a smooth closed oriented manifold. Let $\alpha \in H_2(X;\mathbb{Z})$ be a class in homology,
then we know that $\alpha$ can be represented by smooth embedded surfaces. Let $g_X : H_2(X ;\mathbb{Z} ) \rightarrow \mathbb{Z}_{\geq 0}$
assign a homology class the minimum genus over all surfaces representing that class. Notice that $g_X(\alpha) = g_X(-\alpha)$ and $g_X = g_{\overline{X}}$.
Exercise: if $f$ is a diffeomorphism of $X$, then $g_X(f_*(\alpha)) = g_X(\alpha)$. Follows by looking at the image of the genus minimiziing surface under the diffeomorphism.
If $X$ is symplectic and $C \subset X$ is a complex curve, then we have the following adjunction equality
$$
\chi(C) = 2g(C) -2 = [C]^2 - \left< c_1(X), [C]\right>
$$
where $c_1(X)$ is the first Chern class of the tangent bundle of $X$ which can be given a complex structure as $X$ is symplectic. Moreover, $J$
can be chosen so that $C\subset X$ is J-holomorphic.
Now there are remarkably few 4-manifolds for which we completely understand the genus function. Of course we know that $S^4$ has
no interesting $H_2$ so the genus function for the zero class is simply zero.
__Theorem__ _(Kronheimer Mrowka) In $H_2(\mathbb{CP}^2;\mathbb{Z}) = \left< h\right>$, $d \neq 0$, then $g_{\mathbb{CP}^2}(d\cdot h) = \frac{1}{2}(|d| - 1)(|d| - 2)$._
This theorem is known as the Thom conjecture and the proof of Kronheimer and Mrowka used Seiberg-Witten theory.
Now let $X = S^2 \times S^2$. If $\alpha = ax + by$ with $ a, b \geq 0$, then we know that if $a, b > 0$ then $g(\alpha) = (a-1)(b-1)$ and if either
$a$ or $b$ are zero then $g(\alpha = 0)$. Now it is not so hard to find surfaces which realize these genera, but how do we check minimality? The idea is to use Seiberg-Witten invariants!
__Pseudo-Definition__ The Seiberg Witten invariant $SW_X : H^2(X; \mathbb{Z}) \rightarrow \mathbb{Z}$ counts solutions of a certain set of PDEs associated to a class $\alpha \in H^2(X)$.
__Theorem (Adjunction inequality)__ _If $\Sigma \subset X$ is a surface such that $ \Sigma \cdot \Sigma \geq 0$, $g(\Sigma) > 0$ and
$SW_X(K) \neq 0$ then $2 g(\Sigma) - 2 \geq \Sigma \cdot \Sigma + |\left< K , [\Sigma]\right>|$ _
__Theorem__ (Taubes) _If $X$ is symplectic and $b_2^+(X) > 1$ then $SW_X(c_1(X)) = \pm 1$. _
__Theorem (Ozsvàth and Szabò)__ _Let $(X,\omega)$ be symplectic and $C \subset X$ a symplectic submanifold and $\Sigma\subset X$ an embedded surface. If
$[\Sigma] = [C]$ then $g(\Sigma) \geq g(C)$._
Now
Lecture 4:
__Theorem__ $g_{2\mathbb{CP}^2}(6,2) = 10$.
_Proof Sketch_ Suppose that $\Sigma \subset 2 \mathbb{CP}^2$ represents $(6,2)$ and $g(\Sigma) = 9$. Take the double branched cover of $X$. We know that
this is spin? $|H_1(X)| < \infty$ and $\chi(X) = 2 \chi(2\mathbb{CP}^2) - (2-2g) = 2 g + 6$. If $X$ is spin then $b_2^+ = 1$ implies $b_2 = 2$. $b_2^{+} = 2$ implies that $b_2 = 4$.
We also obtain a $\mathbb{Z}_2$ action which also plays a role?
What about a genus function on $K3$? What are the self intersections of embedded spheres in $K3$?
We can show that $[S]^2 \leq 0$.
Fact:
References
========================================================================
Books
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[#Milnorstashef68] Milnor and Stasheff, Characteristic Classes, 1968.
[#GompfStipsicz99] Robert Gompf and Andras Stipsicz, 4-Manifolds Kirby Calculus, 1999.
[#LawsonMichelsohn89] H. Blaine Lawson and Marie-Louise Michelsohn, Princeton University Press, 1989.
Papers
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