K-theory proof of index theorem - some minor confusion

Question

I am trying to understand the general approach to the $K$-theory proof of the Atiyah-Singer index theorem, using this https://arxiv.org/pdf/math/0504555.pdf paper. I ran into some confusion on page 29, where the following is stated:

"It only remains to show that the analytic index commutes with the Thom isomorphism $\phi:K(X)\to K(V)$ where $V$ is a complex vector bundle over $X$. [...] This problem is considerably simplified if we consider trivial bundles which can be expressed as the product $V = X \times\mathbb{R}^n$."

On the same page, it goes on to consider a vector bundle $Y$ which seems to be the associated bundle of some principal $G$-bundle, but the author again considers $P\times_{O(n)} \mathbb{R}^n$, that is, a real vector bundle. I don't quite understand how this makes sense, if we want to prove something for complex vector bundles. I get that we can view a complex vector bundle as a real vector bundle by just "forgetting" about the complex structure, but since the Thom isomorphism (at least in the paper) is only defined for complex vector bundles, I think I am missing something more important. I cannot quite put my finger on it, so if someone could explain the construction on page 29, that would be greatly appreciated.

Answer 1

It seems that this construction was made for real vector bundles because every complex vector bundle can be regarded as a real vector bundle when discarding the complex structure. I am having some trouble justifying this, since we need to add the complex structure again for the Thom isomorphism, and I would like to hear why we don't use $U(n)$-vector bundles instead, since $U(n)$ is also a compact Lie group. Can we not form any complex vector bundle in this way, like we can form any real vector bundle as the associated bundle of some principal bundle?

Answer 2

Recall that if $X$ and $Y$ be are compact smooth manifolds and $i\colon X\hookrightarrow Y$ and is a smooth embedding, we want to define a "shriek map":

$$i_!\colon K_c(TX)\to K_c(TY),$$ where $K_c$ is $K$-theory with compact supports.

The first step (cf. p. 16 of the G. Landweber's article or pp. 497-8 of the original M. Atiyah and I. Singer's The Index of Elliptic Operators: I) is to take take a tubular neighborhood $N\subseteq Y$ of $X$. You can identify it with the normal bundle $N\to X$, which is of course a real vector bundle over $X$. Now observe that $Ti\colon TX\to TY$ is an embedding and that $TN$ is the tubular neighborhood of $TX$. In other words: $TN\to TX$ is a real vector bundle.

But we can say even more. It turns out, that if $\pi\colon TX\to X$ is the projection, then $TN\simeq \pi^*(N\oplus N)$. As $N\oplus N\to X$ can be treated as a complex vector bundle (namely, $N\otimes_\mathbb R \mathbb C)$, we conclude that $TN\to TX$ can be treated as a complex vector bundle as well. In particular it makes sense to consider the Thom homomorphism $K_c(TX)\to K_c(TN)$.

The excision axiom allows us to define the "analytical index" for $N$ as a map $K_c(TN)\to \mathbb Z$. (Note that this "analytical index" is defined via embeddings into compact manifolds, so its meaning is different than in the compact case). We want to show that this analytical index commutes with the Thom homomorphism defined above. To do that we observe that $N$, as a normal bundle over $X$, can be written as $P\times_{O(n)} \mathbb R^n$, where $P$ is a principal $O(n)$-bundle and $X=P/O(n)$. Then one uses the multiplicative axiom of the analytical index. (This is the most advanced part of the proof and in fact motivates the use of equivariant $K$-theory in this case. However, if $N$ is a trivial bundle, $O(n)$ can be replaced by the trivial group $1$, and the equivariance is not needed. Similarly, for orientable $X$, it suffices to consider the group $SO(n)$, what slightly simplifies the proof).