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I am referring to Page 161 of the book "Analysis and Algebra on Differentiable Manifolds: A Workbook for Students and Teachers" by P.M.Gadea, J.Munoz Masque, Springer.

Given that the Lie group $G = \left\{ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \in GL(2,\mathbb{R}) : a_{11} - a_{22} = a_{12} + a_{21}=0 \right\} $ \ The tangent space at the identity element $e$ is $$ T_eG=\{X \in M(2,\mathbb{R}) : Xf=0, f=a_{11}-a_{22} \space or \space f=a_{12} + a_{21} \} \\ = \left\langle \frac{\partial}{\partial x^{1}_{1}}{|_e} + \frac{\partial}{\partial x^{2}_{2}}{|_e} \space,\space -\frac{\partial}{\partial x^{1}_{2}}{|_e} + \frac{\partial}{\partial x^{2}_{1}}{|_e} \right\rangle$$ $$=\left\langle \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \space , \space \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \right\rangle$$ Hence, the Lie algebra of $G$ is $\mathfrak{g}= \left\{ \begin{pmatrix} a_{11} & -a_{21} \\ a_{21} & a_{22} \end{pmatrix} \in M(2,\mathbb{R}) \right\} $ I can't understand the above 3 lines in finding the tangent space $T_eG$, especially the 1st line where $$Xf=0, f=a_{11}-a_{22} \space or \space f=a_{12} + a_{21} $$ Hope someone can enlighten me by giving a more detailed comprehensive explanation of the above solution. Thank you so much.

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Because the function $f=a_{11}-a_{22}$ or $f=a_{12}+a_{21}$ is constantly $0$ on the submanifold, hence for any tangent vector $X$ (at the identity element or anywhere else), we must have $Xf=0$.

On the other hand, by considering the dimensionality, we have $X(a_{11}-a_{22})=X(a_{12}+a_{21})=0$ posed two independent constraint about $X$, and the submaifold clearly has dimension $2=4-2$, hence the two constraints form a complete independent system to cut out $T_e(G)$ out of $T_e(M_2(\mathbb R))=T_e(GL_2(\mathbb R))$.

That is, any $X$ satisfying $X(a_{11}-a_{22})=X(a_{12}+a_{21})=0$ must belong in $T_e(G)$. In particular, $X_1=\frac{\partial}{\partial x_{11}}-\frac{\partial }{\partial x_{22}}$ and $X_2=\frac{\partial}{\partial x_{12}}+\frac{\partial }{\partial x_{21}}$ are both in $T_e(G)$, and they must span the two dimensional vector space.

Just a user
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  • Thank you for your explanation. May I know how to get 2nd line from the 1st line ? Can I understood it as $$ \frac{\partial}{\partial x^{1}{1}} a{11} + \frac{\partial}{\partial x^{2}{2} } (-a{22}) = 1 - 1 =0 $$ ? Similarly for the other basis vector $$ - \frac{\partial}{\partial x^{1}{2}} a{12} + \frac{\partial}{\partial x^{2}{1} }a{21} = -1 + 1 = 0 ? $$ How to find these two basis vectors from 1st line equation ? – JR makoto Dec 30 '22 at 09:01
  • Technically, we need to solve $(\sum c_{ij}\frac{\partial }{\partial x_{ij}}) f = 0$, though this problem is a bit too trivial for the general method. You're essentially right, but also have to take into account that $\frac{\partial a_{11}}{\partial x_{22}}=0$, etc. – Just a user Dec 30 '22 at 10:40