Mixing
Lie Bracket and Matrix Groups
$begingroup$
The lie bracket appears in manifolds and matrix groups. For manifolds a tangent vector $X$ is $$X(p)=sum_{i=1}^n a_i(p) frac{partial}{partial x_i}$$ where there is the parametrization $mathbf{x}:Usubsetmathbb{R}^nto M$ and $frac{partial}{partial x_i}$ is the basis associated with $mathbf{x}$.
Then the change in a function $f$ in the direction of $X$ is
$$(Xf)(p) = sum_i a_i(p) frac{partial f}{partial x_i}(p)$$
The lie bracket of $[X,Y]$ is $(XY-YX)f$ and that expression amounts to applying the above formula for $(Xf)(p)$ multiple times.
Consider a matrix lie group $G$ with lie algebra $T_1(G)$, for example orthogonal matrices and skew-symmetric matrices. The lie bracket is also $XY-YX$ but this is matrix multiplication. This expression can come from considering $C_s(t)=A(s)B(t)A(s)^{-1}$, $C_s'(0)=A(s)YA(s)^{-1}$, $D(s)=A(s)YA(s)^{-1}$,$D'(0)=XY-YX$. But is there a way this can be understood in the previous manifold description of $XY-YX$ which involves tangent vectors where $XYf$ is applying the dot product of a tangent vector with the gradient of a function and doing this twice whereas in the matrix context this was matrix multiplication. Can this all be reconciled?
lie-groups lie-algebras
asked Jan 20 at 2:56
user782220user782220
38812154
$endgroup$
add a comment |
$begingroup$
The lie bracket appears in manifolds and matrix groups. For manifolds a tangent vector $X$ is $$X(p)=sum_{i=1}^n a_i(p) frac{partial}{partial x_i}$$ where there is the parametrization $mathbf{x}:Usubsetmathbb{R}^nto M$ and $frac{partial}{partial x_i}$ is the basis associated with $mathbf{x}$.
Then the change in a function $f$ in the direction of $X$ is
$$(Xf)(p) = sum_i a_i(p) frac{partial f}{partial x_i}(p)$$
The lie bracket of $[X,Y]$ is $(XY-YX)f$ and that expression amounts to applying the above formula for $(Xf)(p)$ multiple times.
Consider a matrix lie group $G$ with lie algebra $T_1(G)$, for example orthogonal matrices and skew-symmetric matrices. The lie bracket is also $XY-YX$ but this is matrix multiplication. This expression can come from considering $C_s(t)=A(s)B(t)A(s)^{-1}$, $C_s'(0)=A(s)YA(s)^{-1}$, $D(s)=A(s)YA(s)^{-1}$,$D'(0)=XY-YX$. But is there a way this can be understood in the previous manifold description of $XY-YX$ which involves tangent vectors where $XYf$ is applying the dot product of a tangent vector with the gradient of a function and doing this twice whereas in the matrix context this was matrix multiplication. Can this all be reconciled?
lie-groups lie-algebras
asked Jan 20 at 2:56
user782220user782220
38812154
$endgroup$
add a comment |
$begingroup$
The lie bracket appears in manifolds and matrix groups. For manifolds a tangent vector $X$ is $$X(p)=sum_{i=1}^n a_i(p) frac{partial}{partial x_i}$$ where there is the parametrization $mathbf{x}:Usubsetmathbb{R}^nto M$ and $frac{partial}{partial x_i}$ is the basis associated with $mathbf{x}$.
Then the change in a function $f$ in the direction of $X$ is
$$(Xf)(p) = sum_i a_i(p) frac{partial f}{partial x_i}(p)$$
The lie bracket of $[X,Y]$ is $(XY-YX)f$ and that expression amounts to applying the above formula for $(Xf)(p)$ multiple times.
Consider a matrix lie group $G$ with lie algebra $T_1(G)$, for example orthogonal matrices and skew-symmetric matrices. The lie bracket is also $XY-YX$ but this is matrix multiplication. This expression can come from considering $C_s(t)=A(s)B(t)A(s)^{-1}$, $C_s'(0)=A(s)YA(s)^{-1}$, $D(s)=A(s)YA(s)^{-1}$,$D'(0)=XY-YX$. But is there a way this can be understood in the previous manifold description of $XY-YX$ which involves tangent vectors where $XYf$ is applying the dot product of a tangent vector with the gradient of a function and doing this twice whereas in the matrix context this was matrix multiplication. Can this all be reconciled?
lie-groups lie-algebras
asked Jan 20 at 2:56
user782220user782220
38812154
$endgroup$
The lie bracket appears in manifolds and matrix groups. For manifolds a tangent vector $X$ is $$X(p)=sum_{i=1}^n a_i(p) frac{partial}{partial x_i}$$ where there is the parametrization $mathbf{x}:Usubsetmathbb{R}^nto M$ and $frac{partial}{partial x_i}$ is the basis associated with $mathbf{x}$.
Then the change in a function $f$ in the direction of $X$ is
$$(Xf)(p) = sum_i a_i(p) frac{partial f}{partial x_i}(p)$$
The lie bracket of $[X,Y]$ is $(XY-YX)f$ and that expression amounts to applying the above formula for $(Xf)(p)$ multiple times.
Consider a matrix lie group $G$ with lie algebra $T_1(G)$, for example orthogonal matrices and skew-symmetric matrices. The lie bracket is also $XY-YX$ but this is matrix multiplication. This expression can come from considering $C_s(t)=A(s)B(t)A(s)^{-1}$, $C_s'(0)=A(s)YA(s)^{-1}$, $D(s)=A(s)YA(s)^{-1}$,$D'(0)=XY-YX$. But is there a way this can be understood in the previous manifold description of $XY-YX$ which involves tangent vectors where $XYf$ is applying the dot product of a tangent vector with the gradient of a function and doing this twice whereas in the matrix context this was matrix multiplication. Can this all be reconciled?
lie-groups lie-algebras
lie-groups lie-algebras
asked Jan 20 at 2:56
user782220user782220
38812154
asked Jan 20 at 2:56
user782220user782220
38812154
asked Jan 20 at 2:56
user782220user782220
38812154
asked Jan 20 at 2:56
user782220user782220
38812154
asked Jan 20 at 2:56
user782220user782220
38812154
38812154
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Remark that $Gl(n,mathbb{R})$ is an open subset of the vector space $M(n,mathbb{R})$ so for every $gin Gl(n,mathbb{R})$, $T_gGl(n,mathbb{R})=M(n,mathbb{R})$. Let $X$ be an element of $M(n,mathbb{R})$ it defines a vector field on $Gl(n,mathbb{R})$ such that $X(g)=gXin T_gGl(n,mathbb{R})$ where we consider the multiplication of matrices.
For every function $f$ defined on $Gl(n,mathbb{R})$, $(Xf)(g)=df_g(X(g)=df_ggX$, we deduce that $(Y(Xf)(g)=d^2f.X(g).Y(g)+YX(g)$ since the differential of $l_X:grightarrow gX$ is $l_X$ since $l_X$ is linear.
This implies that $[X,Y](f)=[XY-YX]g$ since $d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Remark that $Gl(n,mathbb{R})$ is an open subset of the vector space $M(n,mathbb{R})$ so for every $gin Gl(n,mathbb{R})$, $T_gGl(n,mathbb{R})=M(n,mathbb{R})$. Let $X$ be an element of $M(n,mathbb{R})$ it defines a vector field on $Gl(n,mathbb{R})$ such that $X(g)=gXin T_gGl(n,mathbb{R})$ where we consider the multiplication of matrices.
For every function $f$ defined on $Gl(n,mathbb{R})$, $(Xf)(g)=df_g(X(g)=df_ggX$, we deduce that $(Y(Xf)(g)=d^2f.X(g).Y(g)+YX(g)$ since the differential of $l_X:grightarrow gX$ is $l_X$ since $l_X$ is linear.
This implies that $[X,Y](f)=[XY-YX]g$ since $d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
add a comment |
$begingroup$
Remark that $Gl(n,mathbb{R})$ is an open subset of the vector space $M(n,mathbb{R})$ so for every $gin Gl(n,mathbb{R})$, $T_gGl(n,mathbb{R})=M(n,mathbb{R})$. Let $X$ be an element of $M(n,mathbb{R})$ it defines a vector field on $Gl(n,mathbb{R})$ such that $X(g)=gXin T_gGl(n,mathbb{R})$ where we consider the multiplication of matrices.
For every function $f$ defined on $Gl(n,mathbb{R})$, $(Xf)(g)=df_g(X(g)=df_ggX$, we deduce that $(Y(Xf)(g)=d^2f.X(g).Y(g)+YX(g)$ since the differential of $l_X:grightarrow gX$ is $l_X$ since $l_X$ is linear.
This implies that $[X,Y](f)=[XY-YX]g$ since $d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
add a comment |
$begingroup$
Remark that $Gl(n,mathbb{R})$ is an open subset of the vector space $M(n,mathbb{R})$ so for every $gin Gl(n,mathbb{R})$, $T_gGl(n,mathbb{R})=M(n,mathbb{R})$. Let $X$ be an element of $M(n,mathbb{R})$ it defines a vector field on $Gl(n,mathbb{R})$ such that $X(g)=gXin T_gGl(n,mathbb{R})$ where we consider the multiplication of matrices.
For every function $f$ defined on $Gl(n,mathbb{R})$, $(Xf)(g)=df_g(X(g)=df_ggX$, we deduce that $(Y(Xf)(g)=d^2f.X(g).Y(g)+YX(g)$ since the differential of $l_X:grightarrow gX$ is $l_X$ since $l_X$ is linear.
This implies that $[X,Y](f)=[XY-YX]g$ since $d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
Remark that $Gl(n,mathbb{R})$ is an open subset of the vector space $M(n,mathbb{R})$ so for every $gin Gl(n,mathbb{R})$, $T_gGl(n,mathbb{R})=M(n,mathbb{R})$. Let $X$ be an element of $M(n,mathbb{R})$ it defines a vector field on $Gl(n,mathbb{R})$ such that $X(g)=gXin T_gGl(n,mathbb{R})$ where we consider the multiplication of matrices.
For every function $f$ defined on $Gl(n,mathbb{R})$, $(Xf)(g)=df_g(X(g)=df_ggX$, we deduce that $(Y(Xf)(g)=d^2f.X(g).Y(g)+YX(g)$ since the differential of $l_X:grightarrow gX$ is $l_X$ since $l_X$ is linear.
This implies that $[X,Y](f)=[XY-YX]g$ since $d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
edited Jan 20 at 6:59
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
answered Jan 20 at 6:51
Tsemo AristideTsemo Aristide
58.9k11445
58.9k11445
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
add a comment |
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
what is $df_g$ given that $f:Gl(n,mathbf{R})to mathbf{R}$
$endgroup$
– user782220
Jan 21 at 0:22
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$.
$endgroup$
– Tsemo Aristide
Jan 21 at 0:28
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n to R$
$endgroup$
– user782220
Jan 21 at 1:11
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
$begingroup$
The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $mathbb{R}^n$ at $x$.
$endgroup$
– Tsemo Aristide
Jan 21 at 1:30
add a comment |
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