I came across a question which I think I know the answer to, but I'd like some confirmation, or a counter-example to show me where my reason fails me (as it often does....darn unreliable brain

).

the question is this:

is the subgroup of $\displaystyle \mathrm{GL}_n(\mathbb{R})$, given by:

$\displaystyle H = \{A \in \mathrm{GL}_n(\mathbb{R})\ |\text{ } \exists \text{ continuous } f:[0,1] \to \mathrm{GL}_n(\mathbb{R}) \text{ with } f(0) = A, f(1) = I \}$

normal in $\displaystyle \mathrm{GL}_n(\mathbb{R})$?

the topology used on $\displaystyle \mathrm{GL}_n(\mathbb{R})$ is the standard (metric) topology on $\displaystyle \mathbb{R}^{n^2}$, which I believe induces the frobenius norm on the matrices. this norm is sub-multiplicative, so it seems to me, that if $\displaystyle P \in \mathrm{GL}_n(\mathbb{R})$, that:

$\displaystyle g:[0,1] \to \mathrm{GL}_n(\mathbb{R})$ given by:

$\displaystyle g(t) = Pf(t)P^{-1}$ is the desired homotopy of $\displaystyle PAP^{-1}$ with $\displaystyle I$, which would then prove the normality of $\displaystyle H$.

the fly in the ointment being, that $\displaystyle g$ has to be continuous, which is where the frobenius norm comes in.

suppose that for $\displaystyle t_0 \in [0,1], f(t_0) = B$, and denote $\displaystyle f(t) = B_t$.

if $\displaystyle \epsilon > 0$, then if I choose $\displaystyle \delta > 0$ such that:

$\displaystyle |t - t_0| < \delta \implies |B_t - B| < \frac{\epsilon}{|P||P^{-1}|}$, then

$\displaystyle |t - t_0| < \delta \implies |g(t) - g(t_0)| = |PB_tP^{-1} - PBP^{-1}| = |P(B_t - B)P^{-1}|$

$\displaystyle \leq |B_t - B||P||P^{-1}| < \epsilon$, right?