Euler characteristic of matrix manifolds
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I'm reading through examples of computing Euler characteristic of manifolds. I know how to compute it for generic manifolds like sphere and torus. But what about matrix manifolds? I'd like to know how to compute the Euler characteristic of a matrix group, say $SL_3(mathbb{R})$, for example.
What I know: The definition of Euler characteristic for a manifold $M$, I'm using is $chi(M)=L(Id)$, where $L$ is the Lefschetz number of the identity map on $M$, which is basically the intersection number of the diagonal of the identity with itself. I also know the Poincare-Hopf theorem.
Any help is appreciated. Thanks!
manifolds differential-topology intersection-theory
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
$endgroup$
add a comment |
$begingroup$
I'm reading through examples of computing Euler characteristic of manifolds. I know how to compute it for generic manifolds like sphere and torus. But what about matrix manifolds? I'd like to know how to compute the Euler characteristic of a matrix group, say $SL_3(mathbb{R})$, for example.
What I know: The definition of Euler characteristic for a manifold $M$, I'm using is $chi(M)=L(Id)$, where $L$ is the Lefschetz number of the identity map on $M$, which is basically the intersection number of the diagonal of the identity with itself. I also know the Poincare-Hopf theorem.
Any help is appreciated. Thanks!
manifolds differential-topology intersection-theory
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
$endgroup$
$begingroup$
math.stackexchange.com/questions/13260/…
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– Zircht
Dec 8 '18 at 2:29
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@Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way.
$endgroup$
– tangentbundle
Dec 8 '18 at 2:36
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The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero.
$endgroup$
– Charlie Frohman
Dec 8 '18 at 3:09
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@Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups.
$endgroup$
– Mike Miller
Dec 9 '18 at 1:06
add a comment |
$begingroup$
I'm reading through examples of computing Euler characteristic of manifolds. I know how to compute it for generic manifolds like sphere and torus. But what about matrix manifolds? I'd like to know how to compute the Euler characteristic of a matrix group, say $SL_3(mathbb{R})$, for example.
What I know: The definition of Euler characteristic for a manifold $M$, I'm using is $chi(M)=L(Id)$, where $L$ is the Lefschetz number of the identity map on $M$, which is basically the intersection number of the diagonal of the identity with itself. I also know the Poincare-Hopf theorem.
Any help is appreciated. Thanks!
manifolds differential-topology intersection-theory
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
$endgroup$
I'm reading through examples of computing Euler characteristic of manifolds. I know how to compute it for generic manifolds like sphere and torus. But what about matrix manifolds? I'd like to know how to compute the Euler characteristic of a matrix group, say $SL_3(mathbb{R})$, for example.
What I know: The definition of Euler characteristic for a manifold $M$, I'm using is $chi(M)=L(Id)$, where $L$ is the Lefschetz number of the identity map on $M$, which is basically the intersection number of the diagonal of the identity with itself. I also know the Poincare-Hopf theorem.
Any help is appreciated. Thanks!
manifolds differential-topology intersection-theory
manifolds differential-topology intersection-theory
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
asked Dec 8 '18 at 2:10
tangentbundletangentbundle
410211
410211
$begingroup$
math.stackexchange.com/questions/13260/…
$endgroup$
– Zircht
Dec 8 '18 at 2:29
$begingroup$
@Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way.
$endgroup$
– tangentbundle
Dec 8 '18 at 2:36
$begingroup$
The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero.
$endgroup$
– Charlie Frohman
Dec 8 '18 at 3:09
$begingroup$
@Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups.
$endgroup$
– Mike Miller
Dec 9 '18 at 1:06
add a comment |
$begingroup$
math.stackexchange.com/questions/13260/…
$endgroup$
– Zircht
Dec 8 '18 at 2:29
$begingroup$
@Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way.
$endgroup$
– tangentbundle
Dec 8 '18 at 2:36
$begingroup$
The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero.
$endgroup$
– Charlie Frohman
Dec 8 '18 at 3:09
$begingroup$
@Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups.
$endgroup$
– Mike Miller
Dec 9 '18 at 1:06
$begingroup$
math.stackexchange.com/questions/13260/…
$endgroup$
– Zircht
Dec 8 '18 at 2:29
$begingroup$
math.stackexchange.com/questions/13260/…
$endgroup$
– Zircht
Dec 8 '18 at 2:29
$begingroup$
@Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way.
$endgroup$
– tangentbundle
Dec 8 '18 at 2:36
$begingroup$
@Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way.
$endgroup$
– tangentbundle
Dec 8 '18 at 2:36
$begingroup$
The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero.
$endgroup$
– Charlie Frohman
Dec 8 '18 at 3:09
$begingroup$
The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero.
$endgroup$
– Charlie Frohman
Dec 8 '18 at 3:09
$begingroup$
@Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups.
$endgroup$
– Mike Miller
Dec 9 '18 at 1:06
$begingroup$
@Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups.
$endgroup$
– Mike Miller
Dec 9 '18 at 1:06
add a comment |
1 Answer
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You need to pass to something compact first so that we may apply the Lefschetz fixed point theorem.
1) The Euler characteristic is a homotopy invariant.
2) Every connected Lie group has a compact subgroup that it deformation retracts onto. For $SL_n$ it is $SO(n)$: this is a continuous version of the Gram-Schmidt procedure.
3) Now, and only now, may we apply Lefshcetz: Pick any non-identity elemeny of your connected compact group $G$. Left multiplication $L_g$ by that element is a continuous map with no fixed points, so $L(L_g) = 0$. Picking a path from $g$ to the identity $e$ gives a homotopy between $L_g$ and $L_e = text{Id}$. Because the Lefschetz number is (defined to be!) a homotopy invariant, $L(L_e) = chi(G) = 0$.
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
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$begingroup$
You need to pass to something compact first so that we may apply the Lefschetz fixed point theorem.
1) The Euler characteristic is a homotopy invariant.
2) Every connected Lie group has a compact subgroup that it deformation retracts onto. For $SL_n$ it is $SO(n)$: this is a continuous version of the Gram-Schmidt procedure.
3) Now, and only now, may we apply Lefshcetz: Pick any non-identity elemeny of your connected compact group $G$. Left multiplication $L_g$ by that element is a continuous map with no fixed points, so $L(L_g) = 0$. Picking a path from $g$ to the identity $e$ gives a homotopy between $L_g$ and $L_e = text{Id}$. Because the Lefschetz number is (defined to be!) a homotopy invariant, $L(L_e) = chi(G) = 0$.
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
$endgroup$
add a comment |
$begingroup$
You need to pass to something compact first so that we may apply the Lefschetz fixed point theorem.
1) The Euler characteristic is a homotopy invariant.
2) Every connected Lie group has a compact subgroup that it deformation retracts onto. For $SL_n$ it is $SO(n)$: this is a continuous version of the Gram-Schmidt procedure.
3) Now, and only now, may we apply Lefshcetz: Pick any non-identity elemeny of your connected compact group $G$. Left multiplication $L_g$ by that element is a continuous map with no fixed points, so $L(L_g) = 0$. Picking a path from $g$ to the identity $e$ gives a homotopy between $L_g$ and $L_e = text{Id}$. Because the Lefschetz number is (defined to be!) a homotopy invariant, $L(L_e) = chi(G) = 0$.
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
$endgroup$
add a comment |
$begingroup$
You need to pass to something compact first so that we may apply the Lefschetz fixed point theorem.
1) The Euler characteristic is a homotopy invariant.
2) Every connected Lie group has a compact subgroup that it deformation retracts onto. For $SL_n$ it is $SO(n)$: this is a continuous version of the Gram-Schmidt procedure.
3) Now, and only now, may we apply Lefshcetz: Pick any non-identity elemeny of your connected compact group $G$. Left multiplication $L_g$ by that element is a continuous map with no fixed points, so $L(L_g) = 0$. Picking a path from $g$ to the identity $e$ gives a homotopy between $L_g$ and $L_e = text{Id}$. Because the Lefschetz number is (defined to be!) a homotopy invariant, $L(L_e) = chi(G) = 0$.
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
$endgroup$
You need to pass to something compact first so that we may apply the Lefschetz fixed point theorem.
1) The Euler characteristic is a homotopy invariant.
2) Every connected Lie group has a compact subgroup that it deformation retracts onto. For $SL_n$ it is $SO(n)$: this is a continuous version of the Gram-Schmidt procedure.
3) Now, and only now, may we apply Lefshcetz: Pick any non-identity elemeny of your connected compact group $G$. Left multiplication $L_g$ by that element is a continuous map with no fixed points, so $L(L_g) = 0$. Picking a path from $g$ to the identity $e$ gives a homotopy between $L_g$ and $L_e = text{Id}$. Because the Lefschetz number is (defined to be!) a homotopy invariant, $L(L_e) = chi(G) = 0$.
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
answered Dec 9 '18 at 1:04
Mike MillerMike Miller
36.8k470137
36.8k470137
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/13260/…
$endgroup$
– Zircht
Dec 8 '18 at 2:29
$begingroup$
@Zircht I have no knowledge of Lie theory. Most of what's mentioned there makes no sense to me. I'd like to use the definition I mentioned. There has to be a simpler way.
$endgroup$
– tangentbundle
Dec 8 '18 at 2:36
$begingroup$
The tangent bundle of a Lie group is parallelizable via left invariant vector fields. Hence it is trivial and has Euler number zero.
$endgroup$
– Charlie Frohman
Dec 8 '18 at 3:09
$begingroup$
@Zircht The tools used there are waaaaay too hard for this problem. You need those only if you care about calculating the entire homology groups.
$endgroup$
– Mike Miller
Dec 9 '18 at 1:06