A About computing the tangent space at 1 of certain lie groups

[aalma]

Hello :),

I am wondering of the right and direct method to calculate the following tangent spaces at $1$: $T_ISL_n(R)$, $T_IU(n)$ and $T_ISU(n)$.

Definitions I know:
Given a smooth curve $γ : (− ,) → R^n$ with $γ(0) = x$, a tangent vector $˙γ(0)$ is a vector with components $(˙γ1,..., ˙γn)$.
Two (smooth) curves $γ,γ': (− ,) → M$ with $γ(0) = γ'(0) = x$ are called equivalent if in a certain chart containing x (or in any chart) they define the same tangent vector at $x$.
We define $T_x(M)$ to be the set of equivalence classes of curves as above. We note that $T_x(M)$ is a vector space: Choose a chart $a : U → R^n$ containing x. Then the assignment $γ → d/dt (a ◦ γ)(0)$ defines a bijection from $T_x(M)$ to $R^n$. Different charts define different bijections, but they differ by the Jacobi matrix: if $b : U → R^n$ is another chart, the bijections from $Tx(M)$ to $R^n$ differ by the Jacobi matrix at $a(x)$ of the transformation $b◦a^{−1} : a(U) → b(U)$.
Example: Let $G = SO(n,R)$ and let $γ$ be a curve in $G$ with $γ(0) = 1$. Then we can write $γ(s) = 1 + sA + o(s^2)$ where $A ∈ M_{n× n}$ and we want to describe possible values for $A$. Obviously $γ(s)^t = 1 + sA^t + o(s^2)$ so $1 = γ(s)γ(s)^t = 1 + s(A + A^t) + o(s^2)$ which implies that $A$ is skewsymmetric. In the opposite direction, if $A$ is skew-symmetric, $γ(s) = e^{sA}$ is orthogonal and $e^{sA} = 1 + sA + o(s^2)$. We have proven that $T_1(G)$ in this case is the vector space of skew-symmetric matrices.

Can you please suggest a way to compute such tangent spaces.

 

[fresh_42]

Here is the example $GL(n)$ and $SL(n)$
https://www.physicsforums.com/insig...t-iv/#B-Left-Invariant-Vector-Fields-and-GL-n

 

[aalma]

Here is the example $GL(n)$ and $SL(n)$
https://www.physicsforums.com/insig...t-iv/#B-Left-Invariant-Vector-Fields-and-GL-n

Is there a way to calculate the above tangent spaces in a way similar to the example I added?

 

[fresh_42]

Is there a way to calculate the above tangent spaces in a way similar to the example I added?

You need to define curves $\gamma \, : \,(-1,1) \longrightarrow M$ with $\gamma(0)=x$ where $x=I$ is the identity matrix in your case. Then you expand $\gamma(t)$ into a Taylor series, and $\dot\gamma(0)$ will be your tangent vectors. Next, gather so many curves $\gamma(t)$ such that you cover all $\operatorname{dim}M$ many possible directions to be able to span $T_x(M).$

The procedure in the link does the same, only that it uses the canonical Cartesian coordinate system:
$$
T_x(M)=\operatorname{span}\left\{\sum_{i,j}^{\operatorname{dim}M} \left. a_{ij} \dfrac{\partial }{\partial x_{ij}}\right|_{t=0} \gamma(t)\right\}
$$
and the obvious curves
\begin{align*}
\gamma \, : \,(-1,1) &\longrightarrow M\\
t &\longmapsto \begin{pmatrix}1 &0 &\ldots &0\\ \ldots &x_{ij}&\ldots&0 \\ 0&0&\ldots& 1\end{pmatrix} \in M
\end{align*}
plus the defining equations, e.g. $\det(a_{ij})=1$ and / or $(a_{ij}) + (a_{ij})^*=0.$The next problem is to find all curves. These are $\operatorname{dim}M$ many different ones since $\operatorname{dim}T_x(M)=\operatorname{dim}M.$

In the linked article, I have chosen the easiest ones, still given the Cartesian coordinate system. E.g. a typical element of $M=\operatorname{SL}(2)$ looks like
\begin{align*}
X(t)&=\begin{pmatrix}x_{11}(t)&x_{12}(t)\\ x_{21}(t)&x_{22}(t) \end{pmatrix}\; \\&\wedge \;x_{11}(t)\cdot x_{22}(t)- x_{21}(t)\cdot x_{12}(t)=1\; \\&\wedge \;x_{11}(0)=1\, , \,x_{12}(0)=0\, , \,x_{21}(0)=0\, , \,x_{22}(0)=1
\end{align*}

and a typical curve accordingly
$$
\gamma (t)=X(t)\, , \,\gamma(0)=X(0)=I
$$
Differentiation for those matrices is easier than the full Taylor expansion. We are only interested in the second term evaluated at the identity matrix $I$ anyway. This would be the calculation in equations (26) and (27) for the determinant condition:
$$
\left. \dfrac{d}{dt}\right|_{t=0}\operatorname{det}(\gamma(t))=\left. \dfrac{d}{dt}\right|_{t=0}1=0=\operatorname{trace}\left(a_{ij}\right)
$$

Of course, you can choose your own charts (providing the coordinate system), curves (providing the tangent vectors at $x=I$), and calculate the entire Taylor series, but the Cartesian charts, coordinate functions $t\mapsto (x_{ij}(t))$, and the first derivative $\left(\left. \dfrac{d}{dt}\right|_{t=0}x_{ij}(t)\right)$ are likely the easiest you can get. The crucial condition is $\gamma(t)\in M$ with $\gamma(0)=I.$

 

[aalma]

You need to define curves $\gamma \, : \,(-1,1) \longrightarrow M$ with $\gamma(0)=x$ where $x=I$ is the identity matrix in your case. Then you expand $\gamma(t)$ into a Taylor series, and $\dot\gamma(0)$ will be your tangent vectors. Next, gather so many curves $\gamma(t)$ such that you cover all $\operatorname{dim}M$ many possible directions to be able to span $T_x(M).$

The procedure in the link does the same, only that it uses the canonical Cartesian coordinate system:
$$
T_x(M)=\operatorname{span}\left\{\sum_{i,j}^{\operatorname{dim}M} \left. a_{ij} \dfrac{\partial }{\partial x_{ij}}\right|_{t=0} \gamma(t)\right\}
$$
and the obvious curves
\begin{align*}
\gamma \, : \,(-1,1) &\longrightarrow M\\
t &\longmapsto \begin{pmatrix}1 &0 &\ldots &0\\ \ldots &x_{ij}&\ldots&0 \\ 0&0&\ldots& 1\end{pmatrix} \in M
\end{align*}
plus the defining equations, e.g. $\det(a_{ij})=1$ and / or $(a_{ij}) + (a_{ij})^*=0.$The next problem is to find all curves. These are $\operatorname{dim}M$ many different ones since $\operatorname{dim}T_x(M)=\operatorname{dim}M.$

In the linked article, I have chosen the easiest ones, still given the Cartesian coordinate system. E.g. a typical element of $M=\operatorname{SL}(2)$ looks like
\begin{align*}
X(t)&=\begin{pmatrix}x_{11}(t)&x_{12}(t)\\ x_{21}(t)&x_{22}(t) \end{pmatrix}\; \\&\wedge \;x_{11}(t)\cdot x_{22}(t)- x_{21}(t)\cdot x_{12}(t)=1\; \\&\wedge \;x_{11}(0)=1\, , \,x_{12}(0)=0\, , \,x_{21}(0)=0\, , \,x_{22}(0)=1
\end{align*}

and a typical curve accordingly
$$
\gamma (t)=X(t)\, , \,\gamma(0)=X(0)=I
$$
Differentiation for those matrices is easier than the full Taylor expansion. We are only interested in the second term evaluated at the identity matrix $I$ anyway. This would be the calculation in equations (26) and (27) for the determinant condition:
$$
\left. \dfrac{d}{dt}\right|_{t=0}\operatorname{det}(\gamma(t))=\left. \dfrac{d}{dt}\right|_{t=0}1=0=\operatorname{trace}\left(a_{ij}\right)
$$

Of course, you can choose your own charts (providing the coordinate system), curves (providing the tangent vectors at $x=I$), and calculate the entire Taylor series, but the Cartesian charts, coordinate functions $t\mapsto (x_{ij}(t))$, and the first derivative $\left(\left. \dfrac{d}{dt}\right|_{t=0}x_{ij}(t)\right)$ are likely the easiest you can get. The crucial condition is $\gamma(t)\in M$ with $\gamma(0)=I.$

You need to define curves $\gamma \, : \,(-1,1) \longrightarrow M$ with $\gamma(0)=x$ where $x=I$ is the identity matrix in your case. Then you expand $\gamma(t)$ into a Taylor series, and $\dot\gamma(0)$ will be your tangent vectors. Next, gather so many curves $\gamma(t)$ such that you cover all $\operatorname{dim}M$ many possible directions to be able to span $T_x(M).$

The procedure in the link does the same, only that it uses the canonical Cartesian coordinate system:
$$
T_x(M)=\operatorname{span}\left\{\sum_{i,j}^{\operatorname{dim}M} \left. a_{ij} \dfrac{\partial }{\partial x_{ij}}\right|_{t=0} \gamma(t)\right\}
$$
and the obvious curves
\begin{align*}
\gamma \, : \,(-1,1) &\longrightarrow M\\
t &\longmapsto \begin{pmatrix}1 &0 &\ldots &0\\ \ldots &x_{ij}&\ldots&0 \\ 0&0&\ldots& 1\end{pmatrix} \in M
\end{align*}
plus the defining equations, e.g. $\det(a_{ij})=1$ and / or $(a_{ij}) + (a_{ij})^*=0.$The next problem is to find all curves. These are $\operatorname{dim}M$ many different ones since $\operatorname{dim}T_x(M)=\operatorname{dim}M.$

In the linked article, I have chosen the easiest ones, still given the Cartesian coordinate system. E.g. a typical element of $M=\operatorname{SL}(2)$ looks like
\begin{align*}
X(t)&=\begin{pmatrix}x_{11}(t)&x_{12}(t)\\ x_{21}(t)&x_{22}(t) \end{pmatrix}\; \\&\wedge \;x_{11}(t)\cdot x_{22}(t)- x_{21}(t)\cdot x_{12}(t)=1\; \\&\wedge \;x_{11}(0)=1\, , \,x_{12}(0)=0\, , \,x_{21}(0)=0\, , \,x_{22}(0)=1
\end{align*}

and a typical curve accordingly
$$
\gamma (t)=X(t)\, , \,\gamma(0)=X(0)=I
$$
Differentiation for those matrices is easier than the full Taylor expansion. We are only interested in the second term evaluated at the identity matrix $I$ anyway. This would be the calculation in equations (26) and (27) for the determinant condition:
$$
\left. \dfrac{d}{dt}\right|_{t=0}\operatorname{det}(\gamma(t))=\left. \dfrac{d}{dt}\right|_{t=0}1=0=\operatorname{trace}\left(a_{ij}\right)
$$

Of course, you can choose your own charts (providing the coordinate system), curves (providing the tangent vectors at $x=I$), and calculate the entire Taylor series, but the Cartesian charts, coordinate functions $t\mapsto (x_{ij}(t))$, and the first derivative $\left(\left. \dfrac{d}{dt}\right|_{t=0}x_{ij}(t)\right)$ are likely the easiest you can get. The crucial condition is $\gamma(t)\in M$ with $\gamma(0)=I.$

Thanks. Can you please give more deatils of how to do the calculations with taylor series expansion?