# Non embedding of the Heisenberg group

It is well known that Heisenberg groups cannot be bi-Lipschitz embedded into Euclidean spaces. A standard proof uses the fact that a Lipschitz mapping from a Heisenberg group into a Euclidean space is almost everywhere Pansu differentiable. Do you know proofs of this result that do not use Pansu differentiability? I have heard that there are such proofs, but I do not know where to find them.

EDIT: I like all three answers of rob, Robert Young and YCor so I cannot accept any of them since I would like to accept each of them.

• It is not my business but I would accept the most detailed one. Others would serve well as additional references... Apr 16 '18 at 4:30

It can't even be embedded bilipschitz (or even quasi-isometrically) into a Hilbert space. A proof using basic unitary representation theory was given in a paper of mine with Tessera and Valette (GAFA 2007).

The proof works as follows. Let $u$ be such an quasi-isometric embedding (of a compactly generated locally group $G$ for the moment). Then $f:(x,y)\mapsto \|u(x)-u(y)\|$ is a conditionally negative definite kernel on $G$, with the property that $$c^{-1}|x^{-1}y|-C\le \|f(x,y)\|\le c|x^{-1}y|+C$$ for all $x,y$ and some constants $c>0$ and $C$, and $|\cdot|$ denotes the distance (left-invariant) to 1.

Assuming that $G$ is amenable, in a first step change $f$ to a finite perturbation that is uniformly continuous, average $f$ along left cosets, defining $g(x,y)=\int_{k\in G}f(kx,ky)dm$, where $m$ is an invariant mean; then $g$ is continuous (by some little argument, unnecessary if $G$ is discrete, notably the discrete Heisenberg group), and $g$ is a left-invariant conditionally negative definite kernel: $g(x,y)=h(x^{-1}y)$ for some continuous conditionally negative definite function $h$. So, there is an affine isometric representation of $G$ on a Hilbert space such that $h(x)=\|x\cdot 0\|^2$ for all $x\in G$.

The remainder is some stuff on isometric affine action in Hilbert spaces. An old (and quite easy) result of Guichardet is that if $G$ is nilpotent and $\pi$ a unitary/orthogonal representation with no nonzero invariant vectors, then $\overline{H^1}(G,\pi)=0$. Then a simple argument shows that for $G$ arbitrary and an arbitrary unitary/orthogonal representation, every cocycle that is zero in reduced cohomology, has sublinear growth.

So now for $G$ nilpotent as above, we decompose the representation as invariant $\oplus$ (orthogonal of invariant vectors). This decomposes the cocycle in $b+b'$, where $b'$ grows sublinearly. But $b$ being a cocycle of a trivial representation, it's a homomorphism and in particular is zero on $[G,G]$. Hence if $\overline{[G,G]}$ is non-compact, $b+b'$ cannot grow linearly.

• What theorem? I think neither Heisenberg groups not Lipschitz mappings are not mentioned in your paper. Apr 13 '18 at 21:43
• Corollary 4.6 encompasses this. (link: arxiv.org/abs/math/0509527). It does mention neither Heisenberg nor bilipschitz because in both cases it's more general (property $H_{FD}$, resp. quasi-isometric)
– YCor
Apr 13 '18 at 22:00
• (on the other hand, unlike the references mentioned by Robert Young, it does not provide explicit distortion bounds)
– YCor
Apr 13 '18 at 22:07

Assaf Naor's been involved in a few projects along those lines. Austin, Naor, and Tessera proved nonembeddability into certain superreflexive Banach spaces, including Hilbert space, using representation theory. Lafforgue and Naor improved the bound in the superreflexive case using a semigroup argument.

See

Li, Sean Markov convexity and nonembeddability of the Heisenberg group. Ann. Inst. Fourier (Grenoble) 66 (2016), no. 4, 1615–1651.