I want to find $T_{I_n} \ f$ where the map is $\ f : X \mapsto X^2$ with $X\in SL_n(\mathbb{R})$.

By definition of the linear tangent map : $T_{I_n}\ f : T_{I_n}SL_n (\mathbb{R})\to T_{f(I_n)}SL_n(\mathbb{R})$.

But here notice that $f(I_n)=I_n^2=I_n$ and $\ f : SL_n(\mathbb{R})\to SL_n (\mathbb{R})$ because if $\det (X)=1$ then $\det(X^2)=(\det(X))^2=1$.

Now I want to determine $T_{I_n}SL_n (\mathbb{R})$. So I consider the $\mathcal{C}^{\infty}$-map $g: X \mapsto \det(X)-1$ for $X\in SL_n(\mathbb{R})$.

Using the basis of elementary matrix and differentiability theory I find : for $H \in \mathcal{M}_n(\mathbb{R})$ $Dg(X).H=Tr(X^{-1}H).$ Notice that it's a submanifold of $\mathcal{M}_n(\mathbb{R})$ where the dimension is $n^2-1$ (the codim is $1$ by definition of $g$). Indeed the derivative is surjective (we can chose for instance $H=X\neq 0$ and the derivative does not vanish) so $g$ is a submersion.

Now we link this argument to the tangent space. For $X$ the tangent space is : $\ker Dg(X).H=Tr(X^{-1}H)=0$. So for $I_n$ it is $T_{I_n}SL_n(\mathbb{R})=\{H \in \mathcal{M}_n(\mathbb{R})\ / \ Tr(H)=0\}=\mathfrak{sl}(\mathbb{R})$.

So I have to find : $T_{I_n}\ f : \mathfrak{sl}(\mathbb{R}) \to \mathfrak{sl}(\mathbb{R})$. Now I use the argument of drawing a curve on the manifold. I have to build $\gamma : 0\in L\subset \mathbb{R} \to SL_n (\mathbb{R})$ where $\gamma(0)=I_n \in SL_n(\mathbb{R})$. Then if I chose $\gamma(t)=I_n+tX$, $t\in L$, I get $\gamma'(0)=X\in SL_n(\mathbb{R})$. But is $\gamma(t)\in SL_n(\mathbb{R})$ for all $t\in L$ ? Or maybe I have to use the set $\mathfrak{sl}(\mathbb{R})$ ? I think I'm confused with the domains of maps.

To conclude I have to use the equivalence class of curves such that : $T_{I_n} \ f (\overline{\gamma(t)})=\overline{f(\gamma(t))}$.

Thanks in advance !

  • $\begingroup$ Looks like you are not using the proper notation for the description of the function $f$. It would make more sense to write something like $f:SL_n(\mathbb{R})\to SL_n(\mathbb{R}),\quad X\mapsto X^2$. $\endgroup$ – Amitai Yuval May 3 '17 at 3:33
  • $\begingroup$ @AmitaiYuval I don't understand what you meant ? $\endgroup$ – Maman May 3 '17 at 9:54

You are right about the tangent space, which is the Lie algebra $\mathfrak{sl}_n(\mathbb{R})$ consisting of matrices with vanishing trace. For the differential, you don't really have to write specific paths (though it could perhaps be a good exercise). Rather, you can use the Leibniz rule. Let $v\in\mathfrak{sl}_n(\mathbb{R})$, and let $\gamma:(-\epsilon,\epsilon)\to SL_n(\mathbb{R})$ with $\gamma(0)=id,\dot{\gamma}(0)=v$. Then$$\left.\frac{d}{dt}\right|_{t=0}\gamma(t)^2=\dot{\gamma}(0)\gamma(0)+\gamma(0)\dot{\gamma}(0)=2v.$$

  • $\begingroup$ Thank you for the answer but in my attempt why $\gamma(t) \in SL_n(\mathbb{R})$ ? Because in general $\gamma : 0\in L \subset \mathbb{R} \to M$ where $M$ is a manifold (so here it will be $SL_n(\mathbb{R})$). But I have to use also $\mathfrak{sl}_n(\mathbb{R})$... $\endgroup$ – Maman May 3 '17 at 10:20
  • $\begingroup$ @Maman There are vectors $v\in\mathfrak{sl}_n(\mathbb{R})$ for which the path $t\mapsto id+tv$ is not contained in $SL_n(\mathbb{R})$. $\endgroup$ – Amitai Yuval May 3 '17 at 12:44
  • $\begingroup$ $v \in SL_n (\mathbb{R})$ because $v=\gamma'(0)$ and $\gamma' : 0\in L\subset \mathbb{R} \to SL_n(\mathbb{R})$, no ? $\endgroup$ – Maman May 3 '17 at 16:19
  • $\begingroup$ Then I will probably use the canonical injection $\iota$ to arrive in $\mathcal{M}_n(\mathbb{R})$... $\endgroup$ – Maman May 5 '17 at 11:22

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.