I'm trying to read through "The index of elliptic operators I," and here's what I understand of the structure of the proof: 


1. Define (using purely K-theoretic means) a homomorphism $K\_G(TX) \to R(G)$ where $G$ is a compact Lie group, $X$ a $G$-manifold, $R(G)$ the representation ring, and $K\_G(TX)$ the equivariant (compactly supported) K-theory of the tangent bundle $TX$ of a compact manifold $X$. Call this the topological index.
2. Show that the topological index is characterized by a collection of properties: any natural transformation $K\_G(TX) \to R(G)$ that satisfies an excision-like property, a multiplicative property for fiber bundles, and certain normalization conditions is necessarily the topological index.
3. Define (using analysis) the analytical index $K\_G(TX) \to R(G)$ by taking the index (in the usual sense) of an elliptic (pseudo)differential operator.
4. Show that the analytical index satisfies the relevant conditions in 2 (which involves a few computations), so it must be the topological index.


This is all very nice and tidy, but I'm puzzled: what exactly is the role of $G$? Let's say hypothetically that I'm a non-equivariant person and only care about plain differential operators on plain manifolds. It seems (if I understand correctly) that the only place where the $G$-action is relevant is in the multiplicativity condition on the index. 


The reason is that the multiplicativity property is used to show that if 
$i: X \to Y$ is a closed immersion of manifolds, then the induced map $i\_!: K\_G(TX) \to K\_G(TY)$ (given by the Thom isomorphism and push-forward for an open imbedding) preserves the index. 
It seems that the point is to reduce to the case where $X \to Y$ is the imbedding $i: X \to N$ for $ N$ a vector bundle on $X$ and $i$ the zero section. Now in this case there is a principal $O(n)$-bundle $P \to X$ such that $N$ is obtained via $N = P \times\_{O(n)} \mathbb{R}^n$, and Atiyah-Singer define a multiplication 


$$K(X) \times K\_{O(n)} (\mathbb{R}^n) \to K(N)$$


which, when one takes a specified element of $K\_{O(n)}(\mathbb{R}^n)$, is precisely the Thom isomorphism. 


If this is the only place in which equivariance enters the proof, it seems strange that the proof would depend on it. Is it possible to phrase the argument in this paper non-equivariantly? (Or, is there a high-concept reason equivariance should be important to the proof?)



What is the role of equivariance in the Atiyah-Singer index theorem?
As long as you are only interested in Dirac type operators on oriented manifolds, the equivariant $K$-theory can be kicked out of the proof. This is done by Guentner in ''K-homology and the index theorem'', relying on Higson ''On the bordism invariance of the index''. Essentially, the main steps in Guentners argument are:


1. Given a manifold $M \subset R^{2n+1}$ with a Dirac bundle, consider the boundary of the disc bundle of the normal bundle $S$, which is a hypersurface. The Dirac bundle on $M$ induces one on $S$, and tensoring with the Clifford bundle on the sphere, you obtain a Dirac operator on $S$. The topological indices of the original bundle and the new one are the same, and the analytical indices likewise. This procedure involves a bit of equivariant considerations, but no equivariant $K$-theory. That some kind of symmetry considerations are important should not bother you, because it is the symmetry that allows you to do computations (of kernels and cokernels of operators).
2. The hypersurface $S$ is bordant to $S^{2n}$. Both indices are bordism invariant. To be able to use this, the Mayer-Vietoris-sequence in $K$-theory is needed, because the Dirac bundle on $S$ might not be extendable over the bordism. Here the fact that hypersurfaces are $spin^c$ is essential.


This is all quite similar in spirit to IOE1. Why do Atiyah and Singer need equivariant $K$-theory? The question is why they need $K (\mathbb{R}^n)$, equivariant or nonequivariant, at all. Couldn't they just have taken the fibre product with a specific operator on $\mathbb{R}^n$ of index $1$, which should be possible because the operator is equivariant? This is what they essentially do, but the problem is that the product of two pseudodifferential operators of order $0$ is not uniquely defined, but only up to equivalence. So the twisted product of operators is not well-defined on the operator level. They are forced to consider order $0$ operators because only for order $0$ operators it makes sense to ask that they are equal to the identity outside a compact set. Because Atiyah-Singer consider open manifolds, they need order zero operators, and this why at the passage to the normal bundle they need the $K$-theory of $R^n$, which has to be equivariant if the normal bundle of the manifold is not trivial.


Guentner works with closed manifolds, which avoids these problems. Since he takes differential operators only, the product procedure is straightforward. But to get an operator on the sphere of index $1$, he needs orientability, which is why his proof only applies to oriented manifolds.


