Interpretation of the Lie algebra of a Matrix Lie group
I'm looking for an intuitive explanation of the meaning of the Lie algebra for a matrix Lie group from a differential geometry perspective.
Right now, the procedure I've been following is using the Regular Level Set Theorem and showing that for some surjective function $F : M rightarrow N$, if $c$ is a regular value of $F$ then $F^{-1}(c) := { p in M : F(p) = c}$ is an embedded submanifold and $ Lie(F^{-1}(c)) = T_𝟙 F^{-1}(c) cong Ker(F_{*, 𝟙})$.
I've been able to show that the lie algebra of the orthogonal group:
$$
O(n) = { A in M_{n times n}(mathbb{R}) : AA^T = 𝟙 }
$$
is $Skew(n, mathbb{R}) = { A in M_{n times n}(mathbb{R}) : A^T = - A}$, and that the Lie algebra of the Special Linear group:
$$
SL(n, mathbb{R}) = { A in M_{ntimes n} : det(A) = 1 }
$$
is the set of trace $0$ matricies.
The Lie algebras I'm getting seem completely arbitrary to me and I can't really see what the relation is with the original Lie group!
differential-geometry manifolds lie-groups lie-algebras
asked Dec 10 '18 at 4:52
David Feng
738
add a comment |
I'm looking for an intuitive explanation of the meaning of the Lie algebra for a matrix Lie group from a differential geometry perspective.
Right now, the procedure I've been following is using the Regular Level Set Theorem and showing that for some surjective function $F : M rightarrow N$, if $c$ is a regular value of $F$ then $F^{-1}(c) := { p in M : F(p) = c}$ is an embedded submanifold and $ Lie(F^{-1}(c)) = T_𝟙 F^{-1}(c) cong Ker(F_{*, 𝟙})$.
I've been able to show that the lie algebra of the orthogonal group:
$$
O(n) = { A in M_{n times n}(mathbb{R}) : AA^T = 𝟙 }
$$
is $Skew(n, mathbb{R}) = { A in M_{n times n}(mathbb{R}) : A^T = - A}$, and that the Lie algebra of the Special Linear group:
$$
SL(n, mathbb{R}) = { A in M_{ntimes n} : det(A) = 1 }
$$
is the set of trace $0$ matricies.
The Lie algebras I'm getting seem completely arbitrary to me and I can't really see what the relation is with the original Lie group!
differential-geometry manifolds lie-groups lie-algebras
asked Dec 10 '18 at 4:52
David Feng
738
The Lie algebra of a Lie group is just the tangent space at the identity.
– Lord Shark the Unknown
Dec 10 '18 at 4:58
@LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening
– David Feng
Dec 10 '18 at 5:04
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
– Randall
Dec 10 '18 at 5:21
add a comment |
I'm looking for an intuitive explanation of the meaning of the Lie algebra for a matrix Lie group from a differential geometry perspective.
Right now, the procedure I've been following is using the Regular Level Set Theorem and showing that for some surjective function $F : M rightarrow N$, if $c$ is a regular value of $F$ then $F^{-1}(c) := { p in M : F(p) = c}$ is an embedded submanifold and $ Lie(F^{-1}(c)) = T_𝟙 F^{-1}(c) cong Ker(F_{*, 𝟙})$.
I've been able to show that the lie algebra of the orthogonal group:
$$
O(n) = { A in M_{n times n}(mathbb{R}) : AA^T = 𝟙 }
$$
is $Skew(n, mathbb{R}) = { A in M_{n times n}(mathbb{R}) : A^T = - A}$, and that the Lie algebra of the Special Linear group:
$$
SL(n, mathbb{R}) = { A in M_{ntimes n} : det(A) = 1 }
$$
is the set of trace $0$ matricies.
The Lie algebras I'm getting seem completely arbitrary to me and I can't really see what the relation is with the original Lie group!
differential-geometry manifolds lie-groups lie-algebras
asked Dec 10 '18 at 4:52
David Feng
738
I'm looking for an intuitive explanation of the meaning of the Lie algebra for a matrix Lie group from a differential geometry perspective.
Right now, the procedure I've been following is using the Regular Level Set Theorem and showing that for some surjective function $F : M rightarrow N$, if $c$ is a regular value of $F$ then $F^{-1}(c) := { p in M : F(p) = c}$ is an embedded submanifold and $ Lie(F^{-1}(c)) = T_𝟙 F^{-1}(c) cong Ker(F_{*, 𝟙})$.
I've been able to show that the lie algebra of the orthogonal group:
$$
O(n) = { A in M_{n times n}(mathbb{R}) : AA^T = 𝟙 }
$$
is $Skew(n, mathbb{R}) = { A in M_{n times n}(mathbb{R}) : A^T = - A}$, and that the Lie algebra of the Special Linear group:
$$
SL(n, mathbb{R}) = { A in M_{ntimes n} : det(A) = 1 }
$$
is the set of trace $0$ matricies.
The Lie algebras I'm getting seem completely arbitrary to me and I can't really see what the relation is with the original Lie group!
differential-geometry manifolds lie-groups lie-algebras
differential-geometry manifolds lie-groups lie-algebras
asked Dec 10 '18 at 4:52
David Feng
738
asked Dec 10 '18 at 4:52
David Feng
738
edited Dec 10 '18 at 5:22
asked Dec 10 '18 at 4:52
David Feng
738
asked Dec 10 '18 at 4:52
David Feng
738
asked Dec 10 '18 at 4:52
David Feng
738
738
The Lie algebra of a Lie group is just the tangent space at the identity.
– Lord Shark the Unknown
Dec 10 '18 at 4:58
@LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening
– David Feng
Dec 10 '18 at 5:04
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
– Randall
Dec 10 '18 at 5:21
add a comment |
The Lie algebra of a Lie group is just the tangent space at the identity.
– Lord Shark the Unknown
Dec 10 '18 at 4:58
@LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening
– David Feng
Dec 10 '18 at 5:04
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
– Randall
Dec 10 '18 at 5:21
The Lie algebra of a Lie group is just the tangent space at the identity.
– Lord Shark the Unknown
Dec 10 '18 at 4:58
The Lie algebra of a Lie group is just the tangent space at the identity.
– Lord Shark the Unknown
Dec 10 '18 at 4:58
@LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening
– David Feng
Dec 10 '18 at 5:04
@LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening
– David Feng
Dec 10 '18 at 5:04
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
– Randall
Dec 10 '18 at 5:21
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
– Randall
Dec 10 '18 at 5:21
add a comment |
1 Answer
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Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$exp colon mathfrak{g} to G$$
In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.
For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.
answered Dec 10 '18 at 8:45
maridia
1,06512
3
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
add a comment |
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1 Answer
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1 Answer
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votes
Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$exp colon mathfrak{g} to G$$
In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.
For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.
answered Dec 10 '18 at 8:45
maridia
1,06512
3
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
add a comment |
Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$exp colon mathfrak{g} to G$$
In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.
For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.
answered Dec 10 '18 at 8:45
maridia
1,06512
3
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
add a comment |
Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$exp colon mathfrak{g} to G$$
In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.
For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.
answered Dec 10 '18 at 8:45
maridia
1,06512
Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$exp colon mathfrak{g} to G$$
In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.
For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.
answered Dec 10 '18 at 8:45
maridia
1,06512
answered Dec 10 '18 at 8:45
maridia
1,06512
answered Dec 10 '18 at 8:45
maridia
1,06512
answered Dec 10 '18 at 8:45
maridia
1,06512
1,06512
3
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
add a comment |
3
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
3
3
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
Another example that I found illuminating is the computation of $$tag{1}exp tbegin{bmatrix} 0 & -1 \ 1&0end{bmatrix},$$ using the definition $$exp(A)=I+A+frac12 A^2 +ldots $$ The Lie algebra of $SO(2)$ is the linear span of $begin{bmatrix} 0 & -1 \ 1&0end{bmatrix}$, and indeed (1) magically yields $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
– Giuseppe Negro
Dec 10 '18 at 8:57
add a comment |
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The Lie algebra of a Lie group is just the tangent space at the identity.
– Lord Shark the Unknown
Dec 10 '18 at 4:58
@LordSharktheUnknown Yes, but specifically I want to know how the Lie algebra of the above examples relates to the matrix group they come from. I found them by using the fact that the tangent space at the identity is isomorphic to the Kernel of the pushforward of $F$ at the identity, but just using that definition isn't very enlightening
– David Feng
Dec 10 '18 at 5:04
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
– Randall
Dec 10 '18 at 5:21