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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.

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    $\begingroup$ 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
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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.

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

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    $\begingroup$ 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
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    $\begingroup$ 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

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