Finding the tangent space of a subgroup

Question

My professor set the following question and I have an answer, though would like someone with more experience to cast a critical eye over the details as I don't necessarily trust my result!

Define the following set of $2\times 2$ invertible matrices:

$$S=\left\{ \begin{pmatrix} a & 0 \\ b & a \end{pmatrix} \;:\; a,b \in \mathbb{R} \;\text{with}\; a\neq 0\right\}.$$

Firstly, I can prove that this is a subgroup of $GL_{2}(\mathbb{R})$ by showing that for any $A,B\in S$, $AB\in S$ and $A^{-1}\in S$. The next part of the question then asks to define a suitable curve in $S$ and find $T_{e}S$, the tangent space at the identity. So I defined \begin{eqnarray*} \gamma: I\subset\mathbb{R} &\longrightarrow& S \\ t &\longmapsto& A = \gamma(t) = \begin{pmatrix} e^{at} & 0 \\ bt & e^{at} \end{pmatrix} \end{eqnarray*} With this definition, $\gamma(t)\in S$ and $\gamma(0)=I_{2}$, the $2\times 2$ identity matrix. Then $$ \gamma'(t) = \begin{pmatrix} ae^{at} & 0 \\ b & ae^{at} \end{pmatrix} $$ and $$ \gamma'(0) = \begin{pmatrix} a & 0 \\ b & a \end{pmatrix}. $$ Thus the tangent space is $$ T_{e}S = \{ \gamma'(0) \;:\; \gamma(0)=I_{2} \} = \left\{ \begin{pmatrix} a & 0 \\ b & a \end{pmatrix} \;:\; a,b \in \mathbb{R} \;\text{with}\; a\neq 0\right\} = S. $$

This seems to indicate that the tangent space to $S$ at the identity is $S$! This seems odd to me. Any observations/comments/questions/hints would be most welcome. Thanks.

Answer 1

You "forgot" a few curves and tangent vectors.

$T_eS$ must be a two-dimensional vector space, but $S$ is not (but almost :-), We have, for example, $0\not\in S.$ To remedy this, put $a = 0$ in your curve $\gamma$. Please check that this is perfectly valid. We still have $$ \gamma(t) \in S,\quad \gamma(0) = I_2, $$ and in particular $$\gamma'(t) = \begin{pmatrix} ae^{at} & 0 \\ b & ae^{at} \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ b & 0 \end{pmatrix} = \gamma'(0). $$ So we see that allowing the parameter $a = 0$ for the curves $\gamma$ provides exactly the missing elements of $TeS.$ Now we know that $$ U := \left\{ \begin{pmatrix}u & 0 \\ v & u \end{pmatrix} \ \mid\ u,v \in \mathbb R\right\} \subseteq T_eS, $$ and since $U$ is indeed a real vector space with $\dim U = 2 = \dim S = \dim T_eS$ (please check that), we must have equality. $$ U = T_eS. $$ We note that we have $$ S \subseteq T_eS, $$ but that's not so uncommon. It's also true for $GL_n(\mathbb R):$ $$ GL_n(\mathbb R) \subseteq T_eGL_n(\mathbb R) = \mathbb R^{n\times n}. $$

Answer 2

So let me just summarize what I (think) have learnt:

Let $\alpha,\beta \in \mathbb{R}$ (which includes $\alpha=0$) and define the curves \begin{align*} \gamma_{1}:I\subset \mathbb{R} &\longrightarrow S \qquad &\gamma_{2}:I\subset \mathbb{R} &\longrightarrow S \\ t &\longmapsto \gamma_{1}(t)=\begin{pmatrix} e^{\alpha t} & 0 \\ \beta t & e^{\alpha t} \end{pmatrix} \qquad &t &\longmapsto \gamma_{2}(t)=\begin{pmatrix} -e^{\alpha t} & 0 \\ \beta t & -e^{\alpha t} \end{pmatrix}. \end{align*}

This accounts for all possible curves with values in $S$. However, since $\gamma_{2}(0)\neq I_{2}$, this curve will not be in the tangent space $T_{e}S$. Then since \begin{eqnarray*} \gamma_{1}'(t) &=& \begin{pmatrix} \alpha e^{\alpha t} & 0 \\ \beta & \alpha e^{\alpha t} \end{pmatrix} \qquad\Longrightarrow \qquad \gamma_{1}'(0) = \begin{pmatrix} \alpha & 0 \\ \beta & \alpha \end{pmatrix}, \end{eqnarray*} the tangent space is given by: \begin{eqnarray*} T_{e}S = \left\{\begin{pmatrix} \alpha & 0 \\ \beta & \alpha \end{pmatrix} \;:\; \alpha,\beta \in S \right\}. \end{eqnarray*}