Interpretation of the Lie algebra of a Matrix Lie group












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    The Lie algebra of a Lie group is just the tangent space at the identity.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:58










  • $begingroup$
    @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
    $endgroup$
    – David Feng
    Dec 10 '18 at 5:04












  • $begingroup$
    The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:21
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    The Lie algebra of a Lie group is just the tangent space at the identity.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:58










  • $begingroup$
    @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
    $endgroup$
    – David Feng
    Dec 10 '18 at 5:04












  • $begingroup$
    The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:21














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 5:22







David Feng

















asked Dec 10 '18 at 4:52









David FengDavid Feng

758




758












  • $begingroup$
    The Lie algebra of a Lie group is just the tangent space at the identity.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:58










  • $begingroup$
    @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
    $endgroup$
    – David Feng
    Dec 10 '18 at 5:04












  • $begingroup$
    The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:21


















  • $begingroup$
    The Lie algebra of a Lie group is just the tangent space at the identity.
    $endgroup$
    – Lord Shark the Unknown
    Dec 10 '18 at 4:58










  • $begingroup$
    @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
    $endgroup$
    – David Feng
    Dec 10 '18 at 5:04












  • $begingroup$
    The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
    $endgroup$
    – Randall
    Dec 10 '18 at 5:21
















$begingroup$
The Lie algebra of a Lie group is just the tangent space at the identity.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 4:58




$begingroup$
The Lie algebra of a Lie group is just the tangent space at the identity.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 4:58












$begingroup$
@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
$endgroup$
– David Feng
Dec 10 '18 at 5:04






$begingroup$
@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
$endgroup$
– David Feng
Dec 10 '18 at 5:04














$begingroup$
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
$endgroup$
– Randall
Dec 10 '18 at 5:21




$begingroup$
The relation is by tangent spaces at the identity. I don't think you're going to get a better explanation.
$endgroup$
– Randall
Dec 10 '18 at 5:21










1 Answer
1






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 8:57













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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 8:57


















3












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 8:57
















3












3








3





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 8:45









zoidbergzoidberg

1,065113




1,065113








  • 3




    $begingroup$
    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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 8:57
















  • 3




    $begingroup$
    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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
    $endgroup$
    – Giuseppe Negro
    Dec 10 '18 at 8:57










3




3




$begingroup$
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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
$endgroup$
– Giuseppe Negro
Dec 10 '18 at 8:57






$begingroup$
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 $$begin{bmatrix} cos theta & -sin theta \ sin theta & cos thetaend{bmatrix}.$$
$endgroup$
– Giuseppe Negro
Dec 10 '18 at 8:57




















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