0
$\begingroup$

I'm studying the matrix Lie groups like $GL(n,\mathbb{R}),SL(n,\mathbb{R}),O(n),SO(n),U(n),SU(n)$.

1) compactness

For example, to show the compactness of $O(n)$ one should show that it is i) closed and ii) bounded in $M(n, \mathbb{R})$.

And every proof I can find is exactly the same as this one, which argues that since $O(n)$ is the solution set of $A^TA=I$, that is, $O(n)$ is the inverse image of one point $I$ by the map $A\mapsto A^TA$ which is continuous, so it is closed. and since the rows of $O(n)$ form a orthonormal basis, every element is bounded by $1$.

But I don't understand why the map $A\mapsto A^TA$ is continuous and the row of $O(n)$ form a orthonormal basis.

Anyway from this argument, $O(n),SO(n),U(n),SU(n)$ are all compact, and $GL(n,\mathbb{R}),SL(n,\mathbb{R})$ are not. $SL(n,\mathbb{R})$ is closed by the map $A\mapsto\det A$ but is not bounded, and $GL(n,\mathbb{R})$ is open by the same map.

2) connectedness

by the map $A\mapsto\det A$, $GL(n,\mathbb{R})$ is disconnected and $SL(n,\mathbb{R})$ is connected. and $O(n),SO(n),U(n),SU(n)$ are all connected by the maps $A\mapsto A^TA$ and $A\mapsto A^*A$ provided these maps are continuous.

so the questions are, is my argument correct? and why are the maps $A\mapsto A^TA$ and $A\mapsto A^*A$ continuous? and why do the rows of $O(n)$ form a orthonormal basis(not a orthogonal basis)? and do the rows of $U(n)$ form a orthonormal basis like $O(n)$?

$\endgroup$
0
$\begingroup$

1) The map is continuous beacuse it is polynomial in each entry. For instance, when $n=2$ this is the map$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\mapsto\begin{pmatrix}a^2+c^2&ab+cd\\ab+cd&b^2+dd^2\end{pmatrix}.$$

The rows of $O(n)$ form an orthornormal basis because, given a square matrix $A$, asserting that $A\in O(n)$ (that is, asserting that $A^TA=\operatorname{Id}$) is equivalent to: a) each row has norm $1$; b) any two distinct rows or orthogonal. These two conditions hold together if and only if the rows form an orthonormal basis.

2) Saying that “by the map $A\mapsto\det A$ [...] $SL(n,\mathbb{R})$ is connected” makes no sense. Yes, since $\det\bigl(GL(n,\mathbb{R})\bigr)=\mathbb{R}\setminus\{0\}$ and since this is a disconnected set, it follows that $GL(n,\mathbb{R})$ is disconnected. But the image of $GL(n,\mathbb{R})$ by the map $\det^2$ is connected and you don't deduce from that the $GL(n,\mathbb{R})$ is connected, right? So, your argument is wrong.

$\endgroup$
  • $\begingroup$ I don't understand 2). you say $\det(GL(n,\mathbb{R}))=\mathbb{R}-{0}$ is disconnected hence $GL(n,\mathbb{R})$ is disconnected and $\det^2(GL(n,\mathbb{R}))=\mathbb{R}^+$ is connected but it doesn't imply $GL(n,\mathbb{R})$ is connected. why is that? $\endgroup$ – user159234 Jun 7 '17 at 8:40
  • $\begingroup$ Take any non-empty set $S$, connected or not, and consider the map $f\colon S\longrightarrow\mathbb R$ defined by $f(x)=0$. It is a continuous map and its image is $\{0\}$, which is connected. Therefore, the fact that the image is connected does not imply that the set itself is. $\endgroup$ – José Carlos Santos Jun 7 '17 at 8:50
  • $\begingroup$ then what conditions do I need? since I saw a lot of argument using the fact the image of $\det$ is disconnected to show that $GL(n,\mathbb{R})$ is disconnected. $\endgroup$ – user159234 Jun 7 '17 at 9:02
  • $\begingroup$ But that argument is correct. What you cannot do is to see that the image is connected and to deduce that the domain must be connected too. $\endgroup$ – José Carlos Santos Jun 7 '17 at 9:06
  • $\begingroup$ since continuous map preserves open sets, the preimages of the sets $\det>0$ and $\det<0$ are open in $GL(n,\mathbb{R})$ and their intersection is empty, and moreover their union is $GL(n,\mathbb{R})$, so $GL(n,\mathbb{R})$ is disconnected. is this right? if it is, I guess this argument is not applicable for the rest of matrix groups. $\endgroup$ – user159234 Jun 7 '17 at 9:10

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.