Triviality of complexified tangent bundle of a closed surface The Next CEO of Stack OverflowTrivial tangent bundle and orientabilityNon-(stable)-triviality of the tautological bundlesAbout zeros of vector fields in compact surfacesself-intersection of lagrangian submanifoldExample of closed orientable manifold with a nonzero vector field, but not parallelizableIs it (not) possible for two vector fields on the Klein bottle to be a basis?Stably trivial bundle is trivialOn the Euler Class of a manifoldTriviality of vector bundles over Lie groupsNontrivial flat vector bundle over manifold with amenable fundamental group
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Triviality of complexified tangent bundle of a closed surface
The Next CEO of Stack OverflowTrivial tangent bundle and orientabilityNon-(stable)-triviality of the tautological bundlesAbout zeros of vector fields in compact surfacesself-intersection of lagrangian submanifoldExample of closed orientable manifold with a nonzero vector field, but not parallelizableIs it (not) possible for two vector fields on the Klein bottle to be a basis?Stably trivial bundle is trivialOn the Euler Class of a manifoldTriviality of vector bundles over Lie groupsNontrivial flat vector bundle over manifold with amenable fundamental group
$begingroup$
Does anybody know how to prove the following statement:
The complexified tangent bundle $TSotimesmathbbC$ of a closed surface $S$ is topologically trivial iff the Euler characteristic $chi(S)$ is even.
Note that if $S$ is orientable, then $chi(S)=2-2g(S)$ is always even. And in fact it is easy to see $TSotimesmathbbC$ is trivial in the orientable case because $TS$ is stably trivialized by normal bundle in $mathbbR^3$.
I have trouble dealing with the non-orientable case. I can’t even see an idea for $mathbbRP^2$ or the Klein bottle.
algebraic-topology differential-topology vector-bundles non-orientable-surfaces
edited 2 days ago
Michael Albanese
64.3k1599313
asked Mar 27 at 18:41
YeahYeah
183
$endgroup$
add a comment |
$begingroup$
Does anybody know how to prove the following statement:
The complexified tangent bundle $TSotimesmathbbC$ of a closed surface $S$ is topologically trivial iff the Euler characteristic $chi(S)$ is even.
Note that if $S$ is orientable, then $chi(S)=2-2g(S)$ is always even. And in fact it is easy to see $TSotimesmathbbC$ is trivial in the orientable case because $TS$ is stably trivialized by normal bundle in $mathbbR^3$.
I have trouble dealing with the non-orientable case. I can’t even see an idea for $mathbbRP^2$ or the Klein bottle.
algebraic-topology differential-topology vector-bundles non-orientable-surfaces
edited 2 days ago
Michael Albanese
64.3k1599313
asked Mar 27 at 18:41
YeahYeah
183
$endgroup$
$begingroup$
Maybe I'm being silly, but I don't understand your proof for $S$ orientable.
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese Let $nu$ be normal bundle of $S$ in $mathbbR^3$, then $TSoplus(StimesmathbbR)cong TSoplusnucong StimesmathbbR^3$. Then complexify these bundles, and take first Chern classes.
$endgroup$
– Yeah
2 days ago
$begingroup$
I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TSotimes_mathbbRmathbbC) = c_1(TSoplusoverlineTS) = c_1(TS) + c_1(overlineTS) = c_1(TS) - c_1(TS) = 0.$$
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TLotimesmathbbC$ is equivalent to the existence of a Lagrangian immersion $LtomathbbC^n_st$.
$endgroup$
– Yeah
2 days ago
add a comment |
$begingroup$
Does anybody know how to prove the following statement:
The complexified tangent bundle $TSotimesmathbbC$ of a closed surface $S$ is topologically trivial iff the Euler characteristic $chi(S)$ is even.
Note that if $S$ is orientable, then $chi(S)=2-2g(S)$ is always even. And in fact it is easy to see $TSotimesmathbbC$ is trivial in the orientable case because $TS$ is stably trivialized by normal bundle in $mathbbR^3$.
I have trouble dealing with the non-orientable case. I can’t even see an idea for $mathbbRP^2$ or the Klein bottle.
algebraic-topology differential-topology vector-bundles non-orientable-surfaces
edited 2 days ago
Michael Albanese
64.3k1599313
asked Mar 27 at 18:41
YeahYeah
183
$endgroup$
Does anybody know how to prove the following statement:
The complexified tangent bundle $TSotimesmathbbC$ of a closed surface $S$ is topologically trivial iff the Euler characteristic $chi(S)$ is even.
Note that if $S$ is orientable, then $chi(S)=2-2g(S)$ is always even. And in fact it is easy to see $TSotimesmathbbC$ is trivial in the orientable case because $TS$ is stably trivialized by normal bundle in $mathbbR^3$.
I have trouble dealing with the non-orientable case. I can’t even see an idea for $mathbbRP^2$ or the Klein bottle.
algebraic-topology differential-topology vector-bundles non-orientable-surfaces
algebraic-topology differential-topology vector-bundles non-orientable-surfaces
edited 2 days ago
Michael Albanese
64.3k1599313
asked Mar 27 at 18:41
YeahYeah
183
edited 2 days ago
Michael Albanese
64.3k1599313
asked Mar 27 at 18:41
YeahYeah
183
edited 2 days ago
Michael Albanese
64.3k1599313
edited 2 days ago
Michael Albanese
64.3k1599313
edited 2 days ago
Michael Albanese
64.3k1599313
64.3k1599313
asked Mar 27 at 18:41
YeahYeah
183
asked Mar 27 at 18:41
YeahYeah
183
asked Mar 27 at 18:41
YeahYeah
183
183
$begingroup$
Maybe I'm being silly, but I don't understand your proof for $S$ orientable.
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese Let $nu$ be normal bundle of $S$ in $mathbbR^3$, then $TSoplus(StimesmathbbR)cong TSoplusnucong StimesmathbbR^3$. Then complexify these bundles, and take first Chern classes.
$endgroup$
– Yeah
2 days ago
$begingroup$
I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TSotimes_mathbbRmathbbC) = c_1(TSoplusoverlineTS) = c_1(TS) + c_1(overlineTS) = c_1(TS) - c_1(TS) = 0.$$
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TLotimesmathbbC$ is equivalent to the existence of a Lagrangian immersion $LtomathbbC^n_st$.
$endgroup$
– Yeah
2 days ago
add a comment |
$begingroup$
Maybe I'm being silly, but I don't understand your proof for $S$ orientable.
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese Let $nu$ be normal bundle of $S$ in $mathbbR^3$, then $TSoplus(StimesmathbbR)cong TSoplusnucong StimesmathbbR^3$. Then complexify these bundles, and take first Chern classes.
$endgroup$
– Yeah
2 days ago
$begingroup$
I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TSotimes_mathbbRmathbbC) = c_1(TSoplusoverlineTS) = c_1(TS) + c_1(overlineTS) = c_1(TS) - c_1(TS) = 0.$$
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TLotimesmathbbC$ is equivalent to the existence of a Lagrangian immersion $LtomathbbC^n_st$.
$endgroup$
– Yeah
2 days ago
$begingroup$
Maybe I'm being silly, but I don't understand your proof for $S$ orientable.
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
Maybe I'm being silly, but I don't understand your proof for $S$ orientable.
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese Let $nu$ be normal bundle of $S$ in $mathbbR^3$, then $TSoplus(StimesmathbbR)cong TSoplusnucong StimesmathbbR^3$. Then complexify these bundles, and take first Chern classes.
$endgroup$
– Yeah
2 days ago
$begingroup$
@MichaelAlbanese Let $nu$ be normal bundle of $S$ in $mathbbR^3$, then $TSoplus(StimesmathbbR)cong TSoplusnucong StimesmathbbR^3$. Then complexify these bundles, and take first Chern classes.
$endgroup$
– Yeah
2 days ago
$begingroup$
I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TSotimes_mathbbRmathbbC) = c_1(TSoplusoverlineTS) = c_1(TS) + c_1(overlineTS) = c_1(TS) - c_1(TS) = 0.$$
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TSotimes_mathbbRmathbbC) = c_1(TSoplusoverlineTS) = c_1(TS) + c_1(overlineTS) = c_1(TS) - c_1(TS) = 0.$$
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TLotimesmathbbC$ is equivalent to the existence of a Lagrangian immersion $LtomathbbC^n_st$.
$endgroup$
– Yeah
2 days ago
$begingroup$
@MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TLotimesmathbbC$ is equivalent to the existence of a Lagrangian immersion $LtomathbbC^n_st$.
$endgroup$
– Yeah
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $E to X$ is a complex vector bundle of rank $k$, and $X$ is a CW complex of dimension $n < 2k$, then $E cong Foplusvarepsilon_mathbbC^1$ where $F$ is a complex vector bundle of rank $k - 1$. In particular, as $TSotimes_mathbbRmathbbC$ has rank $2$, and $S$ has dimension $2 < 4$, we see that $TSotimesmathbbC cong Loplusvarepsilon_mathbbC^1$ where $L$ is a complex line bundle. As $L$ is determined up to isomorphism by its first Chern class, $TSotimes_mathbbRmathbbC$ is trivial if and only if $c_1(L) = c_1(TSotimes_mathbbRmathbbC) in H^2(S; mathbbZ)$ is zero.
If $S$ is not orientable, then $H^2(S; mathbbZ) cong mathbbZ_2$ and the reduction modulo $2$ map $H^2(S; mathbbZ) to H^2(S; mathbbZ_2)$ is an isomorphism. Under this map, $c_1(TSotimes_mathbbRmathbbC) mapsto w_2(TSotimes_mathbbRmathbbC)$. Now, as a real bundles, $TSotimes_mathbbRmathbbC cong TSoplus TS$, so
$$w_2(TSotimes_mathbbRmathbbC) = w_2(TSoplus TS) = w_2(TS) + w_1(TS)w_1(TS) + w_2(TS) = w_1(TS)^2.$$
On a closed surface $S$, we have $w_2(TS) = w_1(TS)^2$; this is really the statement that the second Wu class $nu_2 = w_1^2 + w_2$ vanishes, see this note for more details. Therefore, we see that $TSotimes_mathbbRmathbbC$ is trivial if and only if $w_2(TS) = 0$. Now note that $langle w_2(TS), [S]rangle equiv chi(S) bmod 2$, so $TSotimes_mathbbRmathbbC$ is trivial if and only if $chi(S)$ is even.
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
add a comment |
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$begingroup$
If $E to X$ is a complex vector bundle of rank $k$, and $X$ is a CW complex of dimension $n < 2k$, then $E cong Foplusvarepsilon_mathbbC^1$ where $F$ is a complex vector bundle of rank $k - 1$. In particular, as $TSotimes_mathbbRmathbbC$ has rank $2$, and $S$ has dimension $2 < 4$, we see that $TSotimesmathbbC cong Loplusvarepsilon_mathbbC^1$ where $L$ is a complex line bundle. As $L$ is determined up to isomorphism by its first Chern class, $TSotimes_mathbbRmathbbC$ is trivial if and only if $c_1(L) = c_1(TSotimes_mathbbRmathbbC) in H^2(S; mathbbZ)$ is zero.
If $S$ is not orientable, then $H^2(S; mathbbZ) cong mathbbZ_2$ and the reduction modulo $2$ map $H^2(S; mathbbZ) to H^2(S; mathbbZ_2)$ is an isomorphism. Under this map, $c_1(TSotimes_mathbbRmathbbC) mapsto w_2(TSotimes_mathbbRmathbbC)$. Now, as a real bundles, $TSotimes_mathbbRmathbbC cong TSoplus TS$, so
$$w_2(TSotimes_mathbbRmathbbC) = w_2(TSoplus TS) = w_2(TS) + w_1(TS)w_1(TS) + w_2(TS) = w_1(TS)^2.$$
On a closed surface $S$, we have $w_2(TS) = w_1(TS)^2$; this is really the statement that the second Wu class $nu_2 = w_1^2 + w_2$ vanishes, see this note for more details. Therefore, we see that $TSotimes_mathbbRmathbbC$ is trivial if and only if $w_2(TS) = 0$. Now note that $langle w_2(TS), [S]rangle equiv chi(S) bmod 2$, so $TSotimes_mathbbRmathbbC$ is trivial if and only if $chi(S)$ is even.
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
add a comment |
$begingroup$
If $E to X$ is a complex vector bundle of rank $k$, and $X$ is a CW complex of dimension $n < 2k$, then $E cong Foplusvarepsilon_mathbbC^1$ where $F$ is a complex vector bundle of rank $k - 1$. In particular, as $TSotimes_mathbbRmathbbC$ has rank $2$, and $S$ has dimension $2 < 4$, we see that $TSotimesmathbbC cong Loplusvarepsilon_mathbbC^1$ where $L$ is a complex line bundle. As $L$ is determined up to isomorphism by its first Chern class, $TSotimes_mathbbRmathbbC$ is trivial if and only if $c_1(L) = c_1(TSotimes_mathbbRmathbbC) in H^2(S; mathbbZ)$ is zero.
If $S$ is not orientable, then $H^2(S; mathbbZ) cong mathbbZ_2$ and the reduction modulo $2$ map $H^2(S; mathbbZ) to H^2(S; mathbbZ_2)$ is an isomorphism. Under this map, $c_1(TSotimes_mathbbRmathbbC) mapsto w_2(TSotimes_mathbbRmathbbC)$. Now, as a real bundles, $TSotimes_mathbbRmathbbC cong TSoplus TS$, so
$$w_2(TSotimes_mathbbRmathbbC) = w_2(TSoplus TS) = w_2(TS) + w_1(TS)w_1(TS) + w_2(TS) = w_1(TS)^2.$$
On a closed surface $S$, we have $w_2(TS) = w_1(TS)^2$; this is really the statement that the second Wu class $nu_2 = w_1^2 + w_2$ vanishes, see this note for more details. Therefore, we see that $TSotimes_mathbbRmathbbC$ is trivial if and only if $w_2(TS) = 0$. Now note that $langle w_2(TS), [S]rangle equiv chi(S) bmod 2$, so $TSotimes_mathbbRmathbbC$ is trivial if and only if $chi(S)$ is even.
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
add a comment |
$begingroup$
If $E to X$ is a complex vector bundle of rank $k$, and $X$ is a CW complex of dimension $n < 2k$, then $E cong Foplusvarepsilon_mathbbC^1$ where $F$ is a complex vector bundle of rank $k - 1$. In particular, as $TSotimes_mathbbRmathbbC$ has rank $2$, and $S$ has dimension $2 < 4$, we see that $TSotimesmathbbC cong Loplusvarepsilon_mathbbC^1$ where $L$ is a complex line bundle. As $L$ is determined up to isomorphism by its first Chern class, $TSotimes_mathbbRmathbbC$ is trivial if and only if $c_1(L) = c_1(TSotimes_mathbbRmathbbC) in H^2(S; mathbbZ)$ is zero.
If $S$ is not orientable, then $H^2(S; mathbbZ) cong mathbbZ_2$ and the reduction modulo $2$ map $H^2(S; mathbbZ) to H^2(S; mathbbZ_2)$ is an isomorphism. Under this map, $c_1(TSotimes_mathbbRmathbbC) mapsto w_2(TSotimes_mathbbRmathbbC)$. Now, as a real bundles, $TSotimes_mathbbRmathbbC cong TSoplus TS$, so
$$w_2(TSotimes_mathbbRmathbbC) = w_2(TSoplus TS) = w_2(TS) + w_1(TS)w_1(TS) + w_2(TS) = w_1(TS)^2.$$
On a closed surface $S$, we have $w_2(TS) = w_1(TS)^2$; this is really the statement that the second Wu class $nu_2 = w_1^2 + w_2$ vanishes, see this note for more details. Therefore, we see that $TSotimes_mathbbRmathbbC$ is trivial if and only if $w_2(TS) = 0$. Now note that $langle w_2(TS), [S]rangle equiv chi(S) bmod 2$, so $TSotimes_mathbbRmathbbC$ is trivial if and only if $chi(S)$ is even.
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
$endgroup$
If $E to X$ is a complex vector bundle of rank $k$, and $X$ is a CW complex of dimension $n < 2k$, then $E cong Foplusvarepsilon_mathbbC^1$ where $F$ is a complex vector bundle of rank $k - 1$. In particular, as $TSotimes_mathbbRmathbbC$ has rank $2$, and $S$ has dimension $2 < 4$, we see that $TSotimesmathbbC cong Loplusvarepsilon_mathbbC^1$ where $L$ is a complex line bundle. As $L$ is determined up to isomorphism by its first Chern class, $TSotimes_mathbbRmathbbC$ is trivial if and only if $c_1(L) = c_1(TSotimes_mathbbRmathbbC) in H^2(S; mathbbZ)$ is zero.
If $S$ is not orientable, then $H^2(S; mathbbZ) cong mathbbZ_2$ and the reduction modulo $2$ map $H^2(S; mathbbZ) to H^2(S; mathbbZ_2)$ is an isomorphism. Under this map, $c_1(TSotimes_mathbbRmathbbC) mapsto w_2(TSotimes_mathbbRmathbbC)$. Now, as a real bundles, $TSotimes_mathbbRmathbbC cong TSoplus TS$, so
$$w_2(TSotimes_mathbbRmathbbC) = w_2(TSoplus TS) = w_2(TS) + w_1(TS)w_1(TS) + w_2(TS) = w_1(TS)^2.$$
On a closed surface $S$, we have $w_2(TS) = w_1(TS)^2$; this is really the statement that the second Wu class $nu_2 = w_1^2 + w_2$ vanishes, see this note for more details. Therefore, we see that $TSotimes_mathbbRmathbbC$ is trivial if and only if $w_2(TS) = 0$. Now note that $langle w_2(TS), [S]rangle equiv chi(S) bmod 2$, so $TSotimes_mathbbRmathbbC$ is trivial if and only if $chi(S)$ is even.
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
answered Mar 27 at 20:02
Michael AlbaneseMichael Albanese
64.3k1599313
64.3k1599313
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
add a comment |
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
$begingroup$
Thank you Michael!
$endgroup$
– Yeah
Mar 27 at 20:11
add a comment |
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$begingroup$
Maybe I'm being silly, but I don't understand your proof for $S$ orientable.
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese Let $nu$ be normal bundle of $S$ in $mathbbR^3$, then $TSoplus(StimesmathbbR)cong TSoplusnucong StimesmathbbR^3$. Then complexify these bundles, and take first Chern classes.
$endgroup$
– Yeah
2 days ago
$begingroup$
I see. Alternatively, because $TS$ is a complex bundle, $$c_1(TSotimes_mathbbRmathbbC) = c_1(TSoplusoverlineTS) = c_1(TS) + c_1(overlineTS) = c_1(TS) - c_1(TS) = 0.$$
$endgroup$
– Michael Albanese
2 days ago
$begingroup$
@MichaelAlbanese As a remark, for any smooth $n-$dimensional manifold $L$, the triviality of $TLotimesmathbbC$ is equivalent to the existence of a Lagrangian immersion $LtomathbbC^n_st$.
$endgroup$
– Yeah
2 days ago