I want to find the tangent space $T_A\operatorname{SL}_n(\mathbb R)$, where $A\in\operatorname{SL}_n(\mathbb R)$, which is a $n^2-1$ smooth manifold.
I remember that in the case of a manifold $M$ defined by the equation of a subset $S\subseteq \mathbb R^n$, we can define the tangent space $T_aS$ as $Ker(F_{*a})$, where $F:\mathbb R^n\to\mathbb R$ is a smooth submersion such that $S=F^{-1}(\{c\})$.
If we don't know a method to write explicitly $F_{*a}$ (it's usually the jacobian of the local representative of the map $F$), we can define the t
angent space via specicific curves.
$Ker(F_{*a})=\{v\in T_aS:F_{*a}(v)=0\}$ so at this point I think that the idea is to see the vectors $v\in T_aM$ as the speed of certain curves. In class we have seen that if for example we define a curve $\gamma:(-\varepsilon,\varepsilon)\to M$ s.t. $\gamma(t)=a+tu$, where $u\in T_a M$ then we can find $Ker(F_{*a})$ via the composition $\dot{(F\circ \gamma)}(t)$.
Here is my first doubt: the image of the curve gamma seems to be a neighbourhood of the affine space given by the tangent plane whose direction is given by the possible positions of the vector $u$. Sometimes we write $u$ as a tangent vector, sometimes we write $u$ as an element of the manifold… Why? I know that if $U\subset M$ is an open neighbourhood of the manifold we have $T_aU\cong T_aM$ but how can we now identify a tangent vector (which intuitively has sense if we want to calculate the tangent space) with an element of the manifold?
Now if we look at the exercise. We have seen in class that if we take $\sigma(t)=A+tB$, with $A\in\operatorname{SL}_n(\mathbb R)$ and $B\in\operatorname{M}_n(\mathbb R)$ then we can define the kernel of the pushforward as
$$\dot{(\det\circ \sigma)}(t)=\dfrac{d}{dt}\det(\sigma(t))=\det(\sigma(t))\text{Tr}(\dot\sigma(t)\sigma^{-1}(t))$$
but shouldn't $B$ be an element of the tangent space?
In order to be precise: $\det$ is a map such that $\det:\operatorname{GL}_n(\mathbb R)\to\mathbb R\setminus\{0\}$, hence $\det_{*A}:T_A\operatorname{GL}_n(\mathbb R)\to T_{\det(A)}(\mathbb R\setminus \{0\})$ and the curve $\sigma$ is s.t. $\sigma:(-\varepsilon,\varepsilon)\to \operatorname{GL}_n(\mathbb R)$
Can you please help me understanding the concept of defining the tangent space via the composition of a submersion with a smooth curve?
Thank you in advance to everybody.
Best Answer
It doesn't look like you're expected to study curves inside $\operatorname{SL}_n(\mathbb R)$. In any case, the curve $\sigma(t)=A+tB$ almost certainly leaves the manifold, so it doesn't match the strategy you described.
What you can easily do (and I suspect this is what was done in class) is study the kernel $\ker F_{*A}$ by looking at directional derivatives $F_{*A}(B)$: $$ F_{*A}(B) = \frac{d}{dt}\Big|_{t=0} F(A+tB) = \det(A)\operatorname{Tr}(B A^{-1}). $$ Now we see that $F_{*A}(B) = 0$ exactly when $B=CA$ for some traceless $C \in \operatorname{GL}_n(\mathbb R)$. At the same time, this shows that $F$ is a submersion (as the kernel is $1$-dimensional) and that $T_A F^{-1}(1) = \{ CA : \operatorname{Tr} C = 0 \}$.
If you like, instead of $\sigma(t)=A+tB$ you could use any other curve (in $\operatorname{GL}_n(\mathbb R)$, not in $\operatorname{SL}_n(\mathbb R)$) with the same initial conditions, i.e., $$ \frac{d}{dt}\Big|_{t=0} F(\sigma(t)) = F_{*A}(B) \qquad \text{if } \sigma(0)=A, \ \sigma'(0)=B. $$ But I see no advantage here.