User:TakuyaMurata/Topology notes

Homotopy groups
The theorem is a consequence of the following special case of the homotopy excision theorem.

Indeed, the theorem implies $$p_*: \pi_{q+1}(CS^n, S^n) \to \pi_{q+1} S^{n+1}$$, p a collapsing map, is bijective for and surjective for. The connecting homomorphism $$\delta: \pi_{q+1}(CS^n, S^n) \to \pi_q S^n$$ is bijective for all q and one can easily see the composition $$p_* \circ \delta^{-1}$$ is the same as the suspension homomorphism $$\pi_q S^n \to \pi_{q+1} S^{n+1}$$.

(One has to work a bit for lower terms.)

We note: a principal G-bundle is flat if and only if its structure group admits a reduction to $$G_d$$.

Symbols of elliptic differential operators
Let $$K_c(M)$$ denote the Grothendieck group of the category of complex vector bundles on the one-point compactification of M. It is called the complex K-theory of M with compact support. Explicitly, the elements of $$K_c(M)$$ are stable isomorphism classes of triples $$(E, F, \sigma)$$ where $$\sigma$$ is invertible outside a compact subset of M and triples are stably isomorphic if $$E \otimes P \simeq F \otimes Q$$ for some trivial bundles P and Q on M.

In particular, given the tangent bundle $$\pi: TM \to M$$, any bundle morphism $$\sigma: \pi^* E \to \pi^* F$$ gives a class in $$K_c(TM)$$. By "definition", any representative of a class in $$K_c(TM)$$ is called a symbol of an elliptic operator.

Fredholm operators
Proof: 2. $$\Rightarrow$$ 1.: Let $$x_i$$ be a bounded sequence in $$H_1$$ such that $$Tx_i = 0$$. Then $$(GT - I)x_i = -x_i$$ has a convergent subsequence. Hence, the kernel of T is finite-dimensional. Since $$(TG - I)^* = G^* T^* - I$$ is compact, by the same argument, the kernel of $$T^*$$ is finite-dimensional. Finally, a similar argument shows that the image of T is closed.

1. $$\Rightarrow$$ 2.: We use the decomposition: $$H_1 = (\operatorname{ker}T)^\bot \oplus \operatorname{ker}T, \, H_2 = \operatorname{im}T \oplus (\operatorname{im}T)^\bot.$$ T induces the isomorphism $$(\operatorname{ker}T)^\bot \simeq \operatorname{im}T$$; let $$G'$$ be the inverse of this isomorphism and put
 * $$G = \begin{bmatrix}

G' & \\ & 0 \end{bmatrix}$$. This G then has the desired property since $$(\operatorname{im}T)^\bot = \operatorname{ker}T^*$$ is finite-dimensional. $$\square$$

An operator satisfying the above equivalent conditions is called a Fredholm operator. If T is Fredholm, let
 * $$\operatorname{index} T = \operatorname{dim} \operatorname{ker} T - \operatorname{dim} \operatorname{coker} T= \operatorname{dim} \operatorname{ker} T - \operatorname{dim} \operatorname{ker} T^*$$.

We note that the condition T has closed image follows if $$T^*$$ has finite-dimensional kernel. Indeed, replacing $$H_1$$ by the quotient by the kernel, we can assume T is injective. But then $$T \oplus i$$ is a continuous linear bijection from $$H_1 \oplus W$$ to $$H_2$$, where $$i: W \to H_2$$ is the inclusion of the complementary sunspace to $$\operatorname{im}T$$; whence an isomorphism of Banach spaces. Now, note the image of T is the image of $$H_1 \oplus 0$$ under $$T \oplus i$$.

An index is a special case of an Euler characteristic in the following way. Given a Fredholm operator $$T: H_1 \to H_2$$, consider the complex
 * $$C: 0 \to 0 \to H_1 \overset{T}\to H_2 \to 0 \to 0$$.

Then, with the usual notation $$\chi(C) = \sum (-1)^i \dim H_i(C)$$, we have: $$\chi(C) = \operatorname{index} T.$$ We can use this observation to prove the important logarithm rule of index. Let Fredholm operators $$H_1 \overset{S}\to H_2 \overset{T}\to H_3$$ be given, and associate complexes $$C', C, C''$$ to $$T, TS \oplus I: H_1 \otimes H_2 \to H_3 \otimes H_2, S$$. Then we have the exact sequence of complexes: $$0 \to C' \to C \to C \to 0$$, which gives: $$\chi(C) = \chi(C') + \chi(C)$$; that is,
 * $$\operatorname{index}(TS) = \operatorname{index} T + \operatorname{index} S.$$

The condition 2. says that a Fredholm operator is a precisely a bounded operator that is invertible module the 2-sided ideal of compact operators.

Proof: First assume K is a finite-rank operator; i.e., the image of K has finite-dimension. Using the decomposition $$H = \operatorname{im} K \oplus (H/\operatorname{im} K)$$, we have:
 * $$\operatorname{index}(I - K) = \operatorname{index} (I - K)_{\operatorname{im} K} + \operatorname{index} I_{H/\operatorname{im} K} = 0$$

(since any operator between finite-dimensional vectors spaces has index zero.) For the general case, choose a finite-rank operator $$K_n$$ such that $$\|K - K_n\| < 1$$ (possible since H is a Hilbert space.) Then $$I - (K - K_n)$$ is invertible by geometric series; call the inverse R and we have: $$ R(I - K) + RK_n = I$$. Thus, $$\operatorname{index}(I - K) = \operatorname{index}(R(I - K)) = \operatorname{index}(I - RK_n) = 0.$$ $$\square$$

Proof: Let G be such that $$TG - I, GT - I$$ are compact operators. If S is an operator such that $$\|S\| < \| G \|^{-1}$$, then, by geometric series, $$I - SG$$ is invertible. Then we have $$(I + GS)^{-1}G(T + S) = I + (I + GS)^{-1}(GT - I)$$, which is the identity plus a compact operator. Thus, taking index we get: $$\operatorname{index} G + \operatorname{index}(T + S) = 0$$, while $$\operatorname{index} G = -\operatorname{index} T$$. $$\square$$

Proof: The function $$[0, 1] \to \mathbb{Z}, \, t \mapsto \operatorname{index}(T_t)$$ is constant. $$\square$$

Let T be a linear operator on a Hilbert space. We say T is of trace class if $$\sum |\langle T e_j, f_j \rangle| < \infty$$ for any orthonormal systems $$e_j, f_j$$. For such an operator T, the infinite sum $$\sum \langle T e_j, e_j \rangle$$ is denoted by $$\operatorname{tr} T$$. For any trace class operators T, S, we have: $$\operatorname{tr}(TS) = \sum_j \langle T S e_j, e_j \rangle = \sum_{k, j} \langle T e_k, e_j \rangle \langle S e_j, e_k \rangle$$ which is symmetric in T and S. Thus, $$\operatorname{tr}(TS) = \operatorname{tr}(ST)$$. This implies that $$\operatorname{tr}$$ is independent of the choice of $$e_j$$.

Proof: If $$N = 1$$, then the statement is exactly the invariance of Euler numbers. Indeed, we have: $$T R_1 = R_2 T$$. Then .... For the general case, replace S by $$S' = S(I + R_2 + \cdots + R^{N-1})$$. Then obviously $$TS' = I - R_2^N$$ and, since $$R_2T = T R_1$$, we also have $$S'T = I - R_1^N$$. Thus, the proof reduces to the basic case. $$\square$$

Next, we define an index bundle, which generalizes an index of a Fredholm operator. Let F be the space of Fredholm operators on a separable Hilbert space H and X a compact space. Let also $$T: X \to F$$ be a continuos map. The basic technical issue is that $$\dim \operatorname{ker}T_x$$ need not be constant in x. Hence, we shall use a little trick to handle this issue. Since X is compact, $$\cap_{x \in X} \operatorname{im} T_x$$ has finite codimension. Let $$H_n \subset \cap_{x \in X} \operatorname{im} T_x$$ be a subspace of codimension n and $$p_n$$ be the orthogonal projection onto $$H_n$$. Then we have: $$\operatorname{im} p_n T_x = H_n$$ for any x in X. Since X is compact and index is locally constant, if $$H_n$$ is sufficiently small, then we can have that $$\operatorname{dim} \operatorname{ker} p_n T_x$$ is constant in x. Hence, we can view $$\operatorname{ker} p_n T$$ as a vector bundle (to take care of local triviality, define it as a subbundle of a trivial bundle.) We put:
 * $$\operatorname{index} T = [\operatorname{ker} p_n T_x] - [X \times \mathbb{C}^n]$$

as an element of $$K(X)$$. It is called the index bundle of T . That this is independent of the choice of $$p_n$$ follows from the below:
 * (*) $$\operatorname{ker} p_{n+1} T \simeq \operatorname{ker} p_n T \oplus (X \times \mathbb{C})$$ (i.e., there is a splitting $$\phi$$).

Since $$\operatorname{index}(p_n T) = \operatorname{index}(T)$$, (*) is clear locally. By continuity (dubious?), we can find a finite open cover $$U_i$$ of X and splittings $$\phi_i$$ such that (*) hold on $$U_j$$ with $$\phi_i$$. Let $$\rho_j$$ be a partition of unity subordinate to $$U_j$$ and put $$\theta_i = \sum_j \rho_j (\phi_i - \phi_j)$$. Then $$\phi = \phi_i - \theta_i$$ is independent of i and gives us (*) on X.

It is clear that index is functorial: $$\operatorname{index} (T \circ f) = f^* \operatorname{index} T$$ for any continuos map $$f: X' \to X$$. This implies the homotopy invariance: indeed, with $$i_t: X \to X \times \{t\} \subset X \times [0, 1]$$ and $$T: X \times [0, 1] \to F$$, all continuous, we have:
 * $$\operatorname{index} T_0 = i_0^* \operatorname{index} T = i_1^* \operatorname{index} T = \operatorname{index} T_1$$

where the middle = is by the homotopy invariance of complex K-theory. By the cos/sin rotation argument, this in turn implies $$\operatorname{index}(TS) = \operatorname{index} T + \operatorname{index} S.$$

Proof: The surjectivity of index is not difficult (one uses the fact elements in $$K(X)$$ are of the form $$k - [E]$$ for an integer k and a bundle E) and we skip it. Thus, suppose $$T: X \to F$$ is a continuous map such that $$\operatorname{index} T = 0$$. Then, using (*) with a large n, we have: $$\operatorname{ker} p_n T \simeq X \times \mathbb{C}^n$$ with the splitting $$\phi$$. Then $$p_n T + \phi: X \to B(H)$$ has the image lying in $$B(H)^\times$$. Since $$p_n T$$ and T are homotopic, this shows the exactness at the middle. $$\square$$

Equivariant cohomology
Let M be a manifold and a compact Lie group G act on it as diffeomorphisms. The homotopy quotient $$M/G$$ of M by G is the fiber bundle associated with the universal bundle $$EG$$ and M; that is, $$M/G = EG \times M / \sim, \, (pg, m) \sim (p, gm)$$. The construction gives us the fiber sequence:
 * $$M \to M/G \to BG$$

where M is a fiber over $$BG$$. The cohomology ring $$H^*(M/G; \mathbb{C})$$ is called the equivariant cohomology ring and is denoted by $$H^*_G(M; \mathbb{C})$$. If M is a point, then $$M/G = BG$$ (equality of homotopy type) and so
 * $$H^*_G(\text{pt}) = H^*(BG) \simeq \mathbb{C}[\mathfrak{t}] = \mathbb{C}[x_1, \cdots, x_r]$$.

Here, $$\mathfrak{t}$$ is the complexification of the Lie algebra of a maximal torus T of G and the isomorphism is given by the Weil homomorphism. We note that any equivariant cohomology ring is a graded algebra over the ring $$H^*_G(\text{pt})$$ via the ring homomorphism induced by $$M/G \to BG$$. Similarly, the inclusion $$M \to M/G$$ induces the ring homomorphism $$H_G^*(M) \to H^*(M)$$. If the action of G is trivial, then
 * $$H^*_G(M) = H^*(EG \times_G M) = H^*(EG/G \times M) = H^*(EG/G) \otimes H^*(M)$$.

That is, $$H^*_G(M) = H^*(BG) \otimes H^*(M)$$. The importance of the above is that it applies to $$M^T$$ the set of fixed points.

From now on, we shall only consider the torus case $$G = T$$.