Q:

What is the Atiyah-Singer index theorem about?

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?

A:

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

$\operatorname{Index}(D)$ is the Fredholm index of $D$, a purely analytic datum,
$\operatorname{Index}_{\text{top}}(D)$ 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:

(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.
(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.

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.
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.

