I was just a little bit curious about the general statement of this theorem. Honestly, I am not at all interested in fully understanding this, so it is not that I am too lazy to read plenty of books about it, but I would like to know a little bit more, what this means.

Therefore I would like to go with an example:

Let $(Tf)(x):= \frac{df}{dx}(x)+\sin(x)f(x)$ be a differential operator on $[0,L]$ for some $L \>>0$.

Apparently, the first question would be: Is this operator Fredholm? I do understand what it means for an operator to be Fredholm and I understand the definition of the Fredholm index, but I don't see whether this one actually is such an example of a Fredholm operator. In case that this is true. Where does topology come into play? I know the definition of an index for a path, but this topological index seems to be different. Maybe this example is not that good, as we are not studying something on any abstract manifolds, but still, could anybody elaborate on this?

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    $\begingroup$ I'm not sure this is the best example to illustrate the index theorem, although it may be instructive to think about...On another note, the friendliest example to use as an entry point to the index theorem is probably the Gauss-Bonnet theorem. Are you familiar with that? $\endgroup$ May 31, 2014 at 0:01
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    $\begingroup$ There are some nice comments here by Paul Seigel. $\endgroup$
    – user98602
    May 31, 2014 at 6:48

1 Answer 1


Fredholm theory, originally developed for studying (systems of) differential equations, had been around for several decades when Gel'fand et al., working in the '50s, noticed that the Fredholm index was homotopy invariant, i.e., if $F_t$, $a \leq t \leq b$ is a continuous path of Fredholm operators, then $\operatorname{Index}(F_t)$ is constant in $t$, and so began wondering if the Fredholm index of at least some Fredholm operators could be computed in terms of honest-to-goodness topological invariants. What Atiyah and Singer then showed was that if $D : E \to F$ is an elliptic (pseudo)differential operator between smooth vector bundles $E$ and $F$ over a compact orientable manifold $X$, then $$ \operatorname{Index}(D) = \operatorname{Index}_{\text{top}}(D), $$ where

  1. $\operatorname{Index}(D)$ is the Fredholm index of $D$, a purely analytic datum,
  2. $\operatorname{Index}_{\text{top}}(D)$ is the topological index of $D$, the pairing of a certain cohomology class on $X$, obtained from $D$, with the fundamental homology class $[X]$ of $X$.

Let me now give the two most basic examples of Atiyah--Singer in action:

  1. (Example of odd-dimensional Atiyah--Singer) Let $\gamma : S^1 \to \mathbb{C} \setminus \{0\}$ be a continuous closed path in the plane that doesn't pass through the origin. Recall that the Hardy space on the circle $S^1$ is the closed subspace $$ H^2(S^1) = \left\{ f = \sum_{n=-\infty}^\infty a_n e^{i n x} \in L^2(S^1) \text{ such that } \forall n < 0, \; a_n = 0 \right\} $$ of $L^2(S^1)$, and let $P$ be the orthogonal projection from $L^2(S^1)$ onto $H^2(S^1)$. Let $M_\gamma : L^2(S^1) \to L^2(S^1)$ be the multiplication operator $(M_\gamma f)(x) = \gamma(x)f(x)$, and define the Toeplitz operator $T_\gamma : H^2(S^1) \to H^2(S^1)$ by $T_\gamma = P \circ M_\gamma \circ P$. Then Atiyah-Singer, in this special case, reduces to the Toeplitz index theorem, i.e., $$ \operatorname{Index}(T_\gamma) = -\text{winding number of $\gamma$}, $$ which is certainly a topological invariant of interest.

  2. (Example of even-dimensional Atiyah--Singer) Let $X$ be a Riemann surface, and view $d+d^\ast$ as an operator $(X \times \mathbb{C}) \oplus \wedge^2 T^\ast_{\mathbb{C}} X \to T^\ast_{\mathbb{C}} X$, i.e., as a map from even-degree forms to odd-degree forms. Then, on the one hand, $$ \operatorname{Index}(d+d^\ast) = \chi(X), $$ where $\chi(X)$ is the Euler characteristic of $X$, whilst on the other hand, $$ \operatorname{Index}_{\text{top}}(d+d^\ast) = \frac{1}{2\pi}\int_X K dA, $$ where $K$ is the Gaussian curvature of $X$. Thus, Atiyah--Singer for the elliptic differential operator $d+d^\ast$ boils down to the Gauss--Bonnet theorem, i.e., $$ \int_X K dA = 2\pi\chi(X). $$

Now, I'm afraid your example of $T := \tfrac{d}{dx} + \sin(x)$ isn't really going to tell you all that much. On the one hand, the closed interval $[0,L]$ is a manifold with boundary, and on the other, you need to impose boundary conditions anyway to actually have a well-defined operator, so that you'd might as well take $L = 2k\pi$ for some $k \in \mathbb{N}$ and impose periodic boundary conditions. Then $T$ will define an elliptic differential operator on $L^2(S^1)$ with the same principal symbol $\sigma(T)$ as $D := \tfrac{d}{dx}$, since $$ \sigma(D)(df) := i[D,f] = if^\prime = i[T,f] =: \sigma(T)(df) $$ for all $f \in C^\infty(S^1)$, and hence will have the same topological index. However, you can directly show that $\ker(D) = \mathbb{C}$ and that $\ker(D^\ast) = \ker(-D) = \mathbb{C}$, so that $$ \operatorname{Index}(T) = \operatorname{Index}_{\mathrm{top}}(T) = \operatorname{Index}_{\mathrm{top}}(D) = \operatorname{Index}(D) = 0. $$ Alternatively, since $\sin(x)$ is smooth, the multiplication by $\sin(x)$ defines a bounded operator on the Sobolev spaces $W^{s,2}(S^1)$ for each $s \geq 0$, and hence $T_t := \tfrac{d}{dx} + t\sin(x)$ defines a continuous one-parameter family of elliptic first-order differential operators on $L^2(S^1)$, so that by heat-theoretic methods (cf. Roe, Elliptic operators, topology and asymptotic methods 2nd ed., pp. 144-145), the analytic index $\operatorname{Index}(T_t)$ is constant in $t$, and hence $$ \operatorname{Index}(T) = \operatorname{Index}(T_1) = \operatorname{Index}(T_0) = \operatorname{Index}(D). $$ As it turns out, it's a general fact that the index of an elliptic differential operator on a closed odd-dimensional manifold necessarily vanishes---observe that Toeplitz operators are pseudodifferential, not differential, operators.

Now, what's the point of Atiyah--Singer? The yoga is similar to that of, say, Stokes's theorem, where one side of the equation is sometimes easier to deal with, and the other side sometimes easier instead.

  1. On the one hand, Fredholm indices, which were originally devised for studying systems of differential equations, can be very difficult to compute, but the right-hand-side of Atiyah--Singer is sometimes computable instead.

  2. On the other hand, you might want to know if a certain topological quantity, which is, a priori, only rational, is actually an integer; if you can realise it as the right-hand-side of Atiyah--Singer for some suitable (pseudo)differential operator, then you know that it is an integer, since the left-hand-side is necessarily an integer.

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    $\begingroup$ sorry, now after thinking about this I am only left with one question: I read quite often that if you have a bounded Fredholm operator then this index is invariant under compact perturbations. What you are using there is that if you have a unbounded operator $\frac{d}{dx}$ then this one is invariant under the bounded perturbation $t \sin(x)$, but I just cannot find this theorem. Would you mind giving me a reference? $\endgroup$
    – user66906
    Jan 17, 2015 at 18:08
  • $\begingroup$ The topological index of a differential operator depends only on its principal symbol; since $\tfrac{d}{dx}$ and $\tfrac{d}{dx} + \sin(x)$ have the same highest order term, they therefore have the same principal symbol $\sigma(p) := ip$, and hence the same topological index. $\endgroup$ Jan 17, 2015 at 21:38
  • $\begingroup$ On the other hand, I think you can also show that they have the same analytic index (cf. Roe, Elliptic operators, topology and asymptotic methods 2nd ed., pp. 144-145), though there it isn't enough that multiplication by $\sin(x)$ is just a bounded operator, but that it is multiplication by a smooth function, and thus plays well with respect to the underlying Sobolev theory. $\endgroup$ Jan 17, 2015 at 22:11
  • $\begingroup$ So, on second thought, what I originally wrote, i.e., that $t \mapsto T_t := \tfrac{d}{dx} + t\sin(x)$ is a homotopy of elliptic differential operators, is correct, but you're absolutely right that it's not enough that $T_t$ is a bounded perturbation of $D = T_0$; you need an appeal to Sobolev theory. $\endgroup$ Jan 17, 2015 at 22:21
  • $\begingroup$ Thank you very much. Proposition 11.13 answers all my questions, so far. $\endgroup$
    – user66906
    Jan 17, 2015 at 22:59

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