SL(n;R) is connected

The special feature is that although not every element of $SL(n,\\mathbb{R})$ is in the image of the exponential map of $\\mathfrak{sl}(n,\\mathbb{R})$ , $SL(n,\\mathbb{R})$ is still a connected Lie group. The proof below is a guideline and should be clarified a bit more at some points, but this was done intentionally.

To illustrate the point, first we show

Proposition 0.1.

$\\begin{pmatrix}-1&1\\\\0&-1\\end{pmatrix}\otin\\exp{\\mathfrak{sl}(2,\\mathbb{R})}$ , but it is in $SL(2,\\mathbb{R})$ .

Proof.

$\\det x=:\\det\\begin{pmatrix}-1&1\\\\0&-1\\end{pmatrix}=1$ , so $x\\in SL(2,\\mathbb{R})$ . We see that $x$ is not diagonalizable, it already is in Jordan normal form. Moreover, it has a double eigenvalue, $-1$ . Suppose that $x=\\exp X,X\\in\\mathfrak{sl}(2,\\mathbb{R})$ , then $\\operatorname{tr}{X}=0$ . Since $x$ had a double eigenvalue, so does $X$ , hence the eigenvalues of $X$ both are $0$ . But this implies the eigenvalues of $x$ are $1$ . This is a contradiction.∎

Lemma 0.2.

We have $\\forall x\\in SL(n,\\mathbb{R}):x=\\exp(X_{a})\\exp(X_{s})$ with $X_{a}^{t}=-X_{a},X_{s}^{t}=X_{s}\\in\\mathfrak{sl}(n,\\mathbb{R})$ .

Proof.

The keyword here is polar decomposition.We notice that $x^{t}x$ is symmetric and positive definite, since $\\forall\\psi\\in\\mathbb{R}^{n}:\\langle\\psi,x^{t}x\\psi\\rangle>0$ , with the standard inner product on $\\mathbb{R}^{n}$ . Hence, we can write $x=RP$ , with $P=(x^{t}x)^{\\frac{1}{2}}$ and $R=xP^{-1}$ . $P$ is well defined, since any real symmetric, positive definite matrix is diagonalizable. It’s easy to check that $RR^{t}=\\operatorname{id}_{n}$ , hence $R\\in O(n)$ . We had $\\det{P}>0$ and $\\det{x}=1$ , hence $\\det(R)>0\\Rightarrow\\det{R}=1\\Rightarrow R\\in SO(n\\mathbb{n})$ and so $\\det{P}=1$ . Since the choice of positive root is unique, $R$ and $P$ are unique. Moreover, $SO(n)$ is exactly generated by the set $\\{X\\in GL(n,\\mathbb{R})|X^{t}=-X\\}$ and $\\Omega$ , the set of real symmetric matrices of determinant $1$ , by $\\{X\\in GL(n,\\mathbb{R})|X^{t}=X,\\operatorname{tr}{X}=0\\}$ , we have the wanted statement: $SL(n,\\mathbb{R}\\subset SO(n)\\times\\exp{\\Omega}$ .∎

The reverse inclusion is simply shown: any such combination is trivially in $SL(n,\\mathbb{R})$ .

Corollary 0.3.

$SL(n,\\mathbb{R})$ is connected.

Proof.

This is now clear from the fact that both $SO(n)$ and $\\Omega$ are connected and so $\\forall s,t\\in[0,1]:\\exp{sX}\\exp{tY}\\in SL(n,\\mathbb{R})$ , a fact easily checked by taking the determinant. So $SL(n,\\mathbb{R})$ is path-connected, hence connected.∎

单词	SLnRIsConnected
释义	SL(n;R) is connected The special feature is that although not every element of $SL(n,\\mathbb{R})$ is in the image of the exponential map of $\\mathfrak{sl}(n,\\mathbb{R})$ , $SL(n,\\mathbb{R})$ is still a connected Lie group. The proof below is a guideline and should be clarified a bit more at some points, but this was done intentionally. To illustrate the point, first we show Proposition 0.1. $\\begin{pmatrix}-1&1\\\\0&-1\\end{pmatrix}\otin\\exp{\\mathfrak{sl}(2,\\mathbb{R})}$ , but it is in $SL(2,\\mathbb{R})$ . Proof. $\\det x=:\\det\\begin{pmatrix}-1&1\\\\0&-1\\end{pmatrix}=1$ , so $x\\in SL(2,\\mathbb{R})$ . We see that $x$ is not diagonalizable, it already is in Jordan normal form. Moreover, it has a double eigenvalue, $-1$ . Suppose that $x=\\exp X,X\\in\\mathfrak{sl}(2,\\mathbb{R})$ , then $\\operatorname{tr}{X}=0$ . Since $x$ had a double eigenvalue, so does $X$ , hence the eigenvalues of $X$ both are $0$ . But this implies the eigenvalues of $x$ are $1$ . This is a contradiction.∎ Lemma 0.2. We have $\\forall x\\in SL(n,\\mathbb{R}):x=\\exp(X_{a})\\exp(X_{s})$ with $X_{a}^{t}=-X_{a},X_{s}^{t}=X_{s}\\in\\mathfrak{sl}(n,\\mathbb{R})$ . Proof. The keyword here is polar decomposition.We notice that $x^{t}x$ is symmetric and positive definite, since $\\forall\\psi\\in\\mathbb{R}^{n}:\\langle\\psi,x^{t}x\\psi\\rangle>0$ , with the standard inner product on $\\mathbb{R}^{n}$ . Hence, we can write $x=RP$ , with $P=(x^{t}x)^{\\frac{1}{2}}$ and $R=xP^{-1}$ . $P$ is well defined, since any real symmetric, positive definite matrix is diagonalizable. It’s easy to check that $RR^{t}=\\operatorname{id}_{n}$ , hence $R\\in O(n)$ . We had $\\det{P}>0$ and $\\det{x}=1$ , hence $\\det(R)>0\\Rightarrow\\det{R}=1\\Rightarrow R\\in SO(n\\mathbb{n})$ and so $\\det{P}=1$ . Since the choice of positive root is unique, $R$ and $P$ are unique. Moreover, $SO(n)$ is exactly generated by the set $\\{X\\in GL(n,\\mathbb{R})\|X^{t}=-X\\}$ and $\\Omega$ , the set of real symmetric matrices of determinant $1$ , by $\\{X\\in GL(n,\\mathbb{R})\|X^{t}=X,\\operatorname{tr}{X}=0\\}$ , we have the wanted statement: $SL(n,\\mathbb{R}\\subset SO(n)\\times\\exp{\\Omega}$ .∎ The reverse inclusion is simply shown: any such combination is trivially in $SL(n,\\mathbb{R})$ . Corollary 0.3. $SL(n,\\mathbb{R})$ is connected. Proof. This is now clear from the fact that both $SO(n)$ and $\\Omega$ are connected and so $\\forall s,t\\in[0,1]:\\exp{sX}\\exp{tY}\\in SL(n,\\mathbb{R})$ , a fact easily checked by taking the determinant. So $SL(n,\\mathbb{R})$ is path-connected, hence connected.∎
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