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In the language of $K$-theory, the Atiyah-Singer index theorem says that for a compact manifold $X$ the topological index map $\text{t-index}: K(TX) \to K(T\mathbb R^n) \simeq \mathbb Z$ induced by embedding $X$ in $\mathbb R^n$ is equal to the analytical index map $K(TX) \to \mathbb Z$ obtained by looking at the index of the elliptic operator whose symbol corresponds to the given element in $K(TX)$.

My question is if there is a definition of the topological index map that does not require an embedding into a euclidean space. Clearly by the index theorem we can take the analytic index as a definition, but is there a more topological/geometric intrinsic definition for t-index that is (relatively) easily seen to be equivalent to the above definition?

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I think the best answer you will find is the axiomatic characterization of the topological index in Index of Elliptic Operators I.

One defines an index function to be a map $ind_X: K(TX) \to \mathbb{Z}$ with the properties that $ind_{point}$ is the identity, and for any embedding $i: X \to Y$ the wrong way map $i_!: K(TX) \to K(TY)$ commutes with the index map. It is not hard to show that the index map is actually uniquely characterized by the axioms and that the topological index is an example of an index map. Thus the real content of the index theorem lies in proving that there is an index map which sends the symbol class of an operator to its index.

From this point of view you should think of embedding $X$ in Euclidean space as a computational crutch rather than an essential part of the theorem. If K-theory were invented before De Rham cohomology then we would all be completely satisfied with the topological characterization of the analytic index map explained above because it tells us everything we need to know to compute the analytic index in terms of K-theoretic invariants. But since we tend to prefer cohomology it makes sense to embed our manifold in a space where it is easy to relate K-theory and cohomology.

This all is what I tell myself to calm down before I go to bed at night, but in fact I do not find it totally satisfying. I have no qualms embedding a manifold in Euclidean space when I want to prove the Gauss-Bonnet theorem or the Hirzebruch signature theorem, but I find it rather disturbing that one can prove the Riemann-Roch theorem by starting with an algebraic curve, forgetting all of its beautiful structure, and stuffing it carelessly into Euclidean space. I've always wondered if there is another computational crutch analogous to the topological index which would remember a little more structure.

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  • $\begingroup$ Thanks for this answer! I'll let the question linger a little bit before accepting this. $\endgroup$ Commented May 23, 2011 at 1:20

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