About computing the tangent space at 1 of certain lie groups

In summary: So, in summary, the method to calculate tangent spaces at given points involves defining curves in the manifold that pass through the given point and finding their tangent vectors. These tangent vectors can then be used to span the tangent space at that point. The procedure suggested in the given link uses the canonical Cartesian coordinate system to define the curves and the equations of the manifold to find the tangent vectors. However, different charts and curves can also be used to achieve the same result.
  • #1
aalma
46
1
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.
 
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  • #4
aalma said:
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.##
 
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  • #5
fresh_42 said:
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.##
fresh_42 said:
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?
 

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