Baum Connes conjecture is considered as a far generalisation of the Atiyah Singer index theorem (in K-theoretical formulation). I would like to understand how the latter follows from this conjecture. My guess is that one can consider trivial group $G$ but I don;t know how to proceed.

I find it overwhelmingly difficult to do this question proper justice. So instead I give a highly condensed answer and refer to Connes NCG, Section II.10 for more details (as well as all the literature that has appeared in the meantime concerning the groupoid approach to index theory).

In my view a good way to understand how BC is a vast generalization of Atiyah-Singer index theory, is to study the conjecture in its more general form, as stated for Lie groupoids (i.e. groupoids with a smooth structure).

Connes gives a very slick proof of AS using a deformation groupoid (called the tangent groupoid). A lot of the following needs "unpacking" and further study in order to be useful. I have certainly glossed over a number of subtleties. Let me just try to give a short recapitulation of Connes idea: Let $M$ be a compact manifold without boundary.

## Preparations:

The first insight is that the Fredholm index of a (pseudo-) differential operator can be rewritten in terms of the tangent groupoid: $G_M^t = M \times M \times (0,1] \cup TM \times \{0\}$. The intuition of this object is to glue the groupoid our operator lives on, namely the pair groupoid $M \times M$, to the tangent bundle $TM$ where its principal symbol lives on (viewed as a bundle of additive groups $(T_x M, +)$, hence a groupoid). The units of this groupoid are $M \times [0,1]$. There are various interpretations on how to obtain a natural smooth structure on $G_M^t$ which I won't go into. One way of looking at it is that $TM$ can be viewed as the normal bundle to the embedding of the diagonal $\Delta_M$ inside $M \times M$ and a smooth structure can be transported using the tubular neighborhood theorem (via the exponential mapping).

Consider $C^{\ast}$-algebra of the groupoid and particularly the evaluation at $t = 0$ as $e_0 \colon C^{\ast}(G_M^t) \to C^{\ast}(TM)$ and at $t = 1$ as $e_1 \colon C^{\ast}(G_M^t) \to C^{\ast}(M \times M)$. Then it is not hard to show that in $K$-theory $(e_0)_{\ast} \colon K(C^{\ast}(G_M^t) \to K(C^{\ast}(TM))$ yields an isomorphism (the kernel of $e_0$ is the contractible $C^{\ast}$-algebra $C^{\ast}(M \times M) \otimes C(0,1]$).

Note that $C^{\ast}(TM) \cong C_0(T^{\ast} M)$ via the Fourier transform and $C^{\ast}(M \times M) \cong \mathcal{K}$ are the compact operators on $L^2(M)$.

Now set: $ind_a := (e_1)_{\ast} \colon (e_0)_{\ast}^{-1} \colon K^0(T^{\ast} M) \to K(\mathcal{K}) \cong \mathbb{Z}$.

That $ind_a$ is in fact equal to the Fredholm index follows quickly by a standard device of $C^{\ast}$-algebra $K$-theory (six-term exact sequence).

The second insight is a generalized Thom isomorphism of groupoids which transfers the calculation of the index to a groupoid which is Morita equivalent to a space. Here one needs the embedding of $M$ into $\mathbb{R}^N$.

## Connection to Baum-Connes

The Baum-Connes conjecture for a Lie groupoid $G$ asserts that the assembly mapping $\mu \colon K_0(G) \to K_0(C^{\ast}(G))$ is an isomorphism.
More precisely, what I want to consider here is the *geometric* assembly mapping for Lie groupoids.
This means the map $\mu_{G}^{geo} \colon K_0^{geo}(G) \to K_0(C^{\ast}(G))$.
The precise definition of $K_0^{geo}(G)$ as well as $\mu_{G}^{geo}$ can be found in NCG. I focus here mostly on how they specialise for the *tangent groupoid* defined above.

The connection to BC becomes clear by examining closely the proof of Atiyah-Singer using the tangent groupoid.

**Steps**

Embed $M$ into $\mathbb{R}^N$ and assume that $N$ is *even*.

1) In general $\mu_G^{geo}$ is constructed using a universal $G$-space (for proper $G$ actions) as well as a specially constructed shriek map (using a deformation groupoid associated to the *anchor* of the action of $G$ on its universal space).
Consider $G = G_M^t$, the tangent groupoid from above. Then $\mu_{G_M^t}$ is simply given by the shriek map from a particular action of $G_M^t$ on $Z = M \times [0,1] \times \mathbb{R}^N$. In other words - after some digging through the definitions in NCG - we obtain the well-defined homomorphism
$\mu_{G_M^t} \colon K(C^{\ast}(Z \rtimes G_M^t)) \to K(C^{\ast}(G_M^t))$.

The definition of the space $Z \rtimes G_M^t$ needs some further unpacking: It is actually the orbit space of the action of the groupoid $G_M^t$ on $Z$ which is obtained with the help of a homomorphism $h \colon (G_M^t, \cdot) \to (\mathbb{R}^N, +)$ defined with the help of the embedding $j$ of $M$ into $\mathbb{R}^N$:

$h(x,y,t) = \frac{j(x) - j(y)}{t}$ if $t > 0$ and $h(x,v) = dj(v)$, $v \in T_x M$ if $t= 0$.

2) Key observation: The map $\mu_{G_M^t}$ is the inverse Connes-Thom isomorphism, i.e. the (inverse!) of the equivariant Thom isomorphism: $\mathcal{CT} \colon K(C^{\ast}(G_M^t)) \to K(C^{\ast}(Z \rtimes G_M^t))$.

For the latter isomorphism one observes that $C^{\ast}(Z \rtimes G_M^t) \cong C^{\ast}(G_M^t) \rtimes \mathbb{R}^N$. Here $\mathbb{R}^N$ acts on $C^{\ast}(G_M^t)$ via $\alpha_v(f)(\gamma) = e^{i (v \cdot h(\gamma))} f(\gamma), f \in C_c(G_M^t)$. Using that $N$ is even we apply Bott periodicity to arrive at the isomorphism $\mathcal{CT}$.

3) The Connes-Thom isomorphism is nothing but the inverse geometric assembly map. I have omitted the proof of this (it involves yet another deformation). This is the key tool to prove the Atiyah-Singer index theorem:

Connes calculates the *orbit space* $Z \rtimes G_M^t$ (after defining $h$ with the help of the fixed embedding of $M$ into $\mathbb{R}^N$).
It turns out that $Z \rtimes G_M^t$ is then another deformation of $\mathbb{R}^N$ into the normal bundle $\mathcal{N}$ to the inclusion of $M$ into $\mathbb{R}^N$, i.e. $BG_M^t = (]0,1] \times \mathbb{R}^N) \cup \mathcal{N}$ as a set.
We have the evaluations:
$(e_0^h)_{\ast} \colon K(C^{\ast}(Z \rtimes G_M^t) \to K(\mathcal{N})$ and $(e_1^h)_{\ast} \colon K(C^{\ast}(Z \rtimes G_M^t)) \to K(\mathbb{R}^N)$.

Together with the Thom isomorphism $K^0(TM) \to K^0(\mathcal{N})$ and the Bott isomorphism (contracting $\mathbb{R}^N$ to a point) $K(\mathbb{R}^N) \cong \mathbb{Z}$ we recover the **topological index** $ind_t$ of Atiyah-Singer using $Z \rtimes G_M^t$.

Recall at this point the definition of $ind_a$. The Connes-Thom isomorphism $CT$ (which specializes to the Thom isomorphism for $t = 0$) - or the inverse of $\mu_{G_M^t}$ - provides the last missing piece to obtain equality of $ind_t$ with $ind_a$ proving the Atiyah-Singer index theorem.

**General case**

We can now ask the question: What does the Baum-Connes conjecture tell us about index theory of more general spaces? As observed above, if we consider the tangent groupoid for a compact manifold without boundary, the geometric Baum-Connes assembly map yields basically the Atiyah-Singer theorem. In general the story gets really interesting: We can study manifolds with boundary or more generally foliated manifolds (i.e. manifolds with a family of immersed submanifolds or leaves). There are associated groupoids for such more singular manifolds to which a lot of the machinery of index theory can be adapted. Connes defined potentially useful analogs of the analytical and topological index for the case of foliations. The general analytical index takes values in $K(C^{\ast}(G))$ where $G$ is the holonomy groupoid of the foliation. The Baum-Connes conjecture is in this case indistinguishable from the index problem which states that we ought to find a topological interpretation of this general analytical index (in terms of the $K$-homology of $G$).

requiredto prove Baum-Connes, but they aren't very strong. $\endgroup$ – Paul Siegel Feb 23 '17 at 2:17