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

- 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.
- 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.
- 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.
- 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?)