21
$\begingroup$

Well, I am not getting any hint how to show $GL_n(\mathbb{C})$ is path connected. So far I have thought that let $A$ be any invertible complex matrix and $I$ be the idenity matrix, I was trying to show a path from $A$ to $I$ then define $f(t)=At+(1-t)I$ for $t\in[0,1]$ which is possible continous except where the $\operatorname{det}{f(t)}=0$ i.e. which has $n$ roots and I can choose a path in $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ where $t_1,\dots,t_n$ are roots of $\operatorname{det}{f(t)}=0$, is my thinking was correct? Could anyone tell me the solution?

$\endgroup$
  • 1
    $\begingroup$ This path can fail in liying on $GL_n(\mathbb{C})$. For example, if you take $A=-I$, at the middle point $t=1/2$ you will get $f(t)=0$ $\endgroup$ – matgaio May 1 '12 at 19:45
  • $\begingroup$ +1 Looks good to me. @matgaio, the idea seems to be that you go around that troublesome $t=1/2$ - there is room for that, when you let $t$ be a complex number! It might have been clearer to define $f(t)$ for all complex $t$, exclude the finite set of points, and then select a path from $0$ to $1$. $\endgroup$ – Jyrki Lahtonen May 1 '12 at 19:47
  • $\begingroup$ Uhm, ok. I was understanding he was trying to go straight from $A$ to $I$. $\endgroup$ – matgaio May 1 '12 at 19:50
  • 2
    $\begingroup$ This thread is slightly relevant. $\endgroup$ – Antonio Vargas May 1 '12 at 20:02
30
$\begingroup$
  • If $P$ is a polynomial of degree $n$, the set $\{\lambda,P(\lambda)\neq 0\}$ is path connected (because its complement is finite, so you can pick a polygonal path).
  • Let $P(t):=\det(A+t(I-A))$. We have that $P(0)=\det A\neq 0$, and $P(1)=\det I=1\neq 0$, so we can find a path $\gamma\colon[0,1]\to\mathbb C$ such that $\gamma(0)=0$, $\gamma(1)=1$, and $P(\gamma(t))\neq 0$ for all $t$. Finally, put $\Gamma(t):=A+\gamma(t)(I-A)$.
  • If $B_1$ and $B_2$ are two invertible matrices, consider $\gamma(t):=B_2\cdot\gamma(t)$, where we chose $\gamma$ for $A:=B_2^{-1}B_1$.
$\endgroup$
  • $\begingroup$ Your argument prove that if $f$ is any polynomial (even holomorphic) function on $\bf C^n$, the $f\not=0$ is path -connected. $\endgroup$ – Thomas Feb 21 '16 at 7:14
13
$\begingroup$

Since any matrix $A\in GL_n(\mathbb C)$ has only finitely many eigenvalues, and 0 isn't one of them, there is a point $z\in S^1$ such that the line through the origin containing $z$ doesn't intersect any of the eigenvalues of $A$. Now, consider the path $f(t)=At+z(1-t)I$. This has determinant 0 iff $z(t-1)$ is an eigenvalue of $At$, which happens iff $z(1-1/t)$ is an eigenvalue of $A$ (this doesn't work when $t=0$, but then it is clear that the determinant is non-zero). By construction, it isn't for any $t\in[0,1]$ so this defines a path form $A$ to $zI$. now there is a path not passing through 0 from $z$ to 1, and this gives rise to a path from $zI$ to $I$, and so concatenating the two paths, we get a path from $A$ to $I$, showing that $GL_n(\mathbb C)$ is path connected.

$\endgroup$
11
$\begingroup$

Use $\Gamma(t) = e^{t \log A + (1-t) \log B}$. This is well defined since $A,B$ are invertible. $\Gamma(t)$ is clearly invertible for all $t$, $\Gamma(0) = B$, $\Gamma(1) = A$.

$\endgroup$
  • $\begingroup$ This idea is not mine, I just can't remember where I saw it, it was in a control theory context. $\endgroup$ – copper.hat May 1 '12 at 20:20
  • $\begingroup$ what is $log A$? $\endgroup$ – Thomas Feb 21 '16 at 7:11
  • 1
    $\begingroup$ @Thomas: Any matrix $B$ such that $e^B = A$ will do, to show the existence, pick a branch of $\log$ (call it $l$) that does not pass through any eigenvalue and take a curve $\gamma$ that encircles each eigenvalue once while remaining in the domain of $l$, then let $\log A = {1 \over 2 \pi i} \int_\gamma l(z) (zI-A)^{-1} dz$. $\endgroup$ – copper.hat Feb 21 '16 at 17:28
  • $\begingroup$ Now I remember, a professor of mine, Shankar Sastry, showed this to me. $\endgroup$ – copper.hat Apr 8 '18 at 5:57
  • 1
    $\begingroup$ How would you connect matrix with determinant +1 to one of -1? $\endgroup$ – copper.hat Apr 13 '18 at 14:04
5
$\begingroup$

Following idea shows connectedness of $Gl(n,\mathbb{C})$ ..

As $A\in Gl(n,\mathbb{C})$ you do have an upper triangular matrix which is similar to $A$.

See subgroup of Invertible, Upper triangular matrices as $$\underbrace{\mathbb{C}^*\times\mathbb{C}^*\times \cdots\times\mathbb{C}^*}_{n- times}\times \underbrace{\mathbb{C}\times\mathbb{C}\times \mathbb{C}\times \cdots \times\mathbb{C}}_{\dfrac{n(n-1)}{2}times}$$ As $\mathbb{C}^*\times\mathbb{C}^*\times \cdots\times\mathbb{C}^*$ is connected and $\mathbb{C}$ is connected so is the above product.

See that conjugation is continuous so preserves connectedness.

By which i mean that $\{BUB^{-1} : U - \text{Upper Triangular}\}$ is connected.

$Gl(n,\mathbb{C})$ being union of connected sets is also connected.

See that all these conjugates have Identity matrix in common.

$\endgroup$
  • $\begingroup$ But $\Bbb{C}^*$ is connected. $\endgroup$ – glip-glop Apr 14 '17 at 9:14
  • $\begingroup$ @iota yes, you are correct. I don't know why I have said so before.thanks $\endgroup$ – user87543 Apr 14 '17 at 12:22
4
$\begingroup$

I think the answer by Praphulla Koushik has the (to me) most elementary approach, but it is awkwardly formulated, so I will try to restate it more simply. It suffices to check two statements:

  1. The group $B$ of invertible upper triangular matrices is path-connected.

  2. The union of conjugates of $B$ (each of which contains $I$) equals $GL(n,\Bbb C)$.

Then every $A\in GL(n,\Bbb C)$ can be joined to $I$ by a path that stays within a conjugate of $B$ that contains$~A$.

For point 1. it suffices to observe that $B$ is homeomorphic to the Cartesian product of the group $T$ of invertible diagonal matrices and the set $U$ of strictly upper triangular matrices; the former is a power of $\Bbb C^*$, which is path-connected, and the latter to a power of$~\Bbb C$.

Point 2. follows from the well known fact that any complex matrix is trigonalisable. This is equivalent to saying that every linear operator$~T$ on $\Bbb C^n$ admits a complete flag of $T$-stable subspaces. To find one, one chooses a left-eigenvector (a linear form on$~\Bbb C^n$) of$~T$, whose kernel is a $T$-stable hyperplane$~H$, and applies induction on the dimension to the restriction of $T$ to$~H$ to find the remaining $T$-stable subspaces for the flag.

$\endgroup$
2
$\begingroup$

By singular value decomposition and the fact that every unitary matrix is unitarily diagonalisable, we can write every invertible matrix $A$ as a product of the form $(US_1U^\ast)S_2(VS_3V^\ast)$, where $U,V$ are unitary matrices and $S_1,S_2,S_3$ are invertible diagonal matrices. So, it suffices to show that each such diagonal matrix is path-connected to $I$, and this boils down to the scalar case, which is trivial.

$\endgroup$

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.