What is the "no small subgroups" argument for $GL(n,\mathbb R)$? That is, how do we show that in $GL(n,\mathbb R)$ there exists a neighborhood of the identity which contains no subgroup other than the trivial one? I had some scribbling (for the $n=2$ case) but could not arrive at a clean proof.

  • 7
    $\begingroup$ Just use the fact exponential map from the Lie algebra to the group is diffeomorphism on small neighbourhood of the identity and that $g^{k}=exp(k\cdot x)$ for some $x$ in the Lie algebra... $\endgroup$ – Asaf Aug 2 '12 at 15:21
  • $\begingroup$ @Asaf: I do not see how this suffices without more work. Can you fill in the rest of your argument? $\endgroup$ – Qiaochu Yuan Aug 3 '12 at 17:16

Here is Asaf's agrument expanded a bit. It has the advantage of working for all Lie groups simultaneously.

Given a Lie group $G$ with Lie algebra $\mathfrak{g}$, consider the exponential map $\exp:\mathfrak{g}\rightarrow G$. It is known that it is a diffeomorphism on a small enough open set $U\subseteq\mathfrak{g}$.

Choosing an inner product on $\mathfrak{g}$, we may assume wlog that $U$ has the form $U = \{v\in \mathfrak{g} : \; |v| < \epsilon\}$ for some $\epsilon > 0$. Let $V\subseteq U$ with $V = \{v\in\mathfrak{g} : \; |v| < \epsilon/2\}$.

I claim that $\exp(V)$ contains no nontrivial subgroups. Indeed, suppose $H\subseteq \exp(V)$ is a subgroup and choose $g\in H$ so $g = \exp(v)$ for some $v\in V$. I claim that $2v \in V$ as well. To see this, notice that since $g^2 \in H\subseteq \exp(V)$, we must have $g^2 = \exp(w)$ for some $w\in V$. Then $\exp(w) = g^2 = \exp(v)^2 = \exp(2v)$ which implies $w=2v$ since $\exp|_U$ is a diffeomorphism. Thus, $2v \in V$.

But now can iterate this argument showing $2^n v \in V$ for all $n$. Since $|2^n v| = 2^n |v|$, this implies $v=0$, i.e. that $g =e$ so $H$ is trivial.

  • $\begingroup$ A typo: you are to claim that $2v\in V$, not $2v\in \exp(V)$. $\endgroup$ – Murat Güngör Aug 4 '12 at 12:14

It suffices to show that the powers of some non-identity element $g \in \text{GL}_n(\mathbb{R})$ near the identity "escape from the identity." If $g$ has an eigenvalue not equal to $1$ then this follows by examining eigenvalues (we should take a neighborhood of the identity containing only elements with eigenvalues very close to $1$), so we reduce to the case that $g$ is unipotent. But now we can just compute that $|1 - g^k|$ tends to $\infty$ in, say, Hilbert-Schmidt norm by writing $g$ as an upper-triangular matrix.

Edit: To show that a neighborhood containing elements with eigenvalues very close to $1$ exists, consider the neighborhood of elements whose characteristic polynomial is close to $(\lambda - 1)^n$ (we will be more precise about this). Write $z = \lambda - 1$, so we are trying to show that a polynomial of the form

$$z^n = a_{n-1} z^{n-1} + ... + a_0$$

has small roots if the $a_i$ are chosen to be small. Writing this as $1 = \frac{a_{n-1}}{z} + ... + \frac{a_0}{z^n}$ we have

$$1 \le (|a_{n-1}| + ... + |a_0|) \text{max} \left( \frac{1}{|z|}, \frac{1}{|z|^n} \right)$$

by the triangle inequality. We conclude that if we stipulate $|a_{n-1}| + ... + |a_0| < \text{min}(\epsilon, \epsilon^n)$ then $|z| < \epsilon$.

  • 2
    $\begingroup$ Alternately, the closure of a small subgroup is compact, and a compact subgroup can be conjugated into $\text{O}(n)$ so has no unipotent elements. But this requires the existence of Haar measure... $\endgroup$ – Qiaochu Yuan Aug 2 '12 at 14:36
  • $\begingroup$ Still alternatively, every nontrivial subgroup of $O(n)$ with compact closure has an element whose trace is at most $n-3/2$ (indeed, take a nonidentity element, take some power having eigenvalue $e^{ix}$ with $x\in [2\pi/3,4\pi/3]$). $\endgroup$ – YCor Aug 2 '12 at 17:25
  • $\begingroup$ Qiaochu, you say that 1 has a neighborhood consisting of matrices having eigenvalues close to 1; do you have a simple proof of this? $\endgroup$ – Murat Güngör Aug 3 '12 at 14:57
  • $\begingroup$ @Murat: I have edited in a proof. $\endgroup$ – Qiaochu Yuan Aug 3 '12 at 15:59
  • 1
    $\begingroup$ It is more than a bit high powered, but one can also argue that the topology of "close roots" on the space of polynomials of at most a fixed degree makes it a finite-dimensional topological vector space over a complete field, and there's only one of those in each dimension (and for each field); so that it must also be the same as the topology of "close coefficients". $\endgroup$ – LSpice Mar 24 '16 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.