5
$\begingroup$

Consider the general linear group $$GL(n,\mathbb R)=\{g\in {\mathbb R}^{n\times n}\mid\det(g)\neq 0\}$$ Prove that the derivative of the function $f=\det:{\mathbb R}^{n\times n}\to\mathbb R$ is given by $$df(g)v=\det(g)\operatorname{trace}(g^{-1}v)$$ for every $g\in GL(n,\mathbb R)$ and every $v\in {\mathbb R}^{n\times n}$. Deduce that the special linear group $$SL(n,\mathbb R):=\{g\in GL(n,\mathbb R)\mid\det(g)=1\}$$ is a smooth submainfold of ${\mathbb R}^{n\times n}$

I have a bunch of questions to ask before I can solve it: Is the dimension of $GL(n,\mathbb R)$ $n^2$? And what is the dimension of the special linear group? What are the basis vectors of the special linear group? Is the general linear space is a metric space? I was trying to compute $df(g)v$, and by definition, $df(g)v=\lim_{t\to 0}\dfrac{\det(g+tv)-\det(g)}{t}$. But how can I expand $ \det(g+tv)$?

$\endgroup$
7
  • $\begingroup$ What do you meant by dimension? $\endgroup$
    – copper.hat
    Commented Dec 17, 2014 at 20:04
  • 1
    $\begingroup$ The general linear group is not a vector space, because it doesn't contain the zero operator. Thus, it does not have a dimension. $\endgroup$
    – Nishant
    Commented Dec 17, 2014 at 20:06
  • 2
    $\begingroup$ @Nishant I think the OP means dimension as a manifold. $\endgroup$ Commented Dec 17, 2014 at 20:07
  • 1
    $\begingroup$ The derivative is straightforward to compute for the matrix $I$. $\endgroup$
    – copper.hat
    Commented Dec 17, 2014 at 20:09
  • 1
    $\begingroup$ For the determinant question, this post is helpful. $\endgroup$ Commented Dec 17, 2014 at 20:11

1 Answer 1

4
$\begingroup$

Some answers/hints:

  1. Yes the dimension of $GL(n,\mathbb{R})$ is $n^2$. In fact it's the open subset of $\mathbb{R}^{n^2}$ on which the determinant (which is continuous) is nonzero.

  2. The special linear group is the subset of the general linear group on which the determinant is one. Neither of these is itself a vector space, so talking about basis vectors doesn't make sense. Here's a hint: If we have $f: M\rightarrow \mathbb{R}$ and consider $N = f^{-1}(\{c\})$, for some $c\in \mathbb{R}$, when is $N$ a manifold? What is its dimension when it is a manifold?

  3. There are various ways to find the derivative of $\det$. Try writing it out in coordinates and using multilinearity of the determinant. (This is easiest when $g=I$, so start there. If you like you can deduce the whole formula from its value at $g=I$ by using the matrix multiplication map $GL_n \times GL_n \rightarrow GL_n$ and the chain rule.)

$\endgroup$
4
  • $\begingroup$ Is $f$ the one appears in my question? If it's not, then is it smooth? $\endgroup$
    – pxc3110
    Commented Dec 17, 2014 at 22:18
  • $\begingroup$ Yes, take $f$ smooth. (The hint is about general such $f$. In your problem $M = GL(n, \mathbb{R})$ and $f = \det$, with $c = 1$ and $N = SL(n, \mathbb{R})$.) $\endgroup$
    – aes
    Commented Dec 17, 2014 at 23:45
  • $\begingroup$ I think $g=I$ is not the easiest case. In contrast, I found it easiest to prove when letting $v=g$. So now I'm trying to show that $tr(g^{-1}v)=\lim_{t\to 0} \frac{\det (I+tg^{-1}v)-1}{t}$ $\endgroup$
    – pxc3110
    Commented Dec 18, 2014 at 13:50
  • $\begingroup$ That's equivalent to the $g = I$ case, with $g^{-1}v$ in place of $v$. $\endgroup$
    – aes
    Commented Dec 18, 2014 at 17:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .