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characteristic classes arising from connections are well-defined
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I am reading an excerpt from Morita's book. The story is that if you have a connection $nabla$ on a vector bundle $E$ over some manifold $M$ and a function $M_n(mathbbR)to mathbbR$ which is a homogeneous polynomial invariant under conjugation, then you obtain a form on $M$ by applying $f$ to the curvature matrix $Omega$. What needs to be shown is that $f(Omega)$ is a closed form, and that $[f(Omega)]$ doesn't depend on the connection you started with. The second part is what I'm having trouble with.
What Morita does is suppose we are given connections $nabla^0$ and $nabla^1$ on $E$. Consider the bundle $EtimesmathbbRto MtimesmathbbR$ and define a connection $tildenabla$ on $Etimes mathbbR$ in the following way. Suppose $s$ is a section of $EtimesmathbbR$ which is independent of $t$ (that is, $s(p,t) = s(p,t')$ for all $t, t'$). Then define $tildenabla_fracpartialpartial t s = 0$ and
$$
tildenabla_X s = (1-t)nabla_X^0 s + tnabla_X^1 s
$$
for $Xin T_(p,t)(MtimesmathbbR)$.
I understand that every vector field on $Mtimes mathbbR$ can be written as a linear combination with function coefficients of $fracpartialpartial t$ and vector fields which are tangent to $Mtimes t$. Morita says that also every section of $Etimes mathbbR$ can be written as a linear combination with function coefficients of sections which are independent of $t$. I don't see why this is true.
Moreover, aren't you supposed to be feeding connections vector fields and not tangent vectors? Is the point here that a vector field on $MtimesmathbbR$ which is tangent to $Mtimes t$ can be considered as a vector field on $M$?
differential-geometry vector-bundles connections characteristic-classes
asked Jul 26 at 23:05
Ross
213
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up vote
2
down vote
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I am reading an excerpt from Morita's book. The story is that if you have a connection $nabla$ on a vector bundle $E$ over some manifold $M$ and a function $M_n(mathbbR)to mathbbR$ which is a homogeneous polynomial invariant under conjugation, then you obtain a form on $M$ by applying $f$ to the curvature matrix $Omega$. What needs to be shown is that $f(Omega)$ is a closed form, and that $[f(Omega)]$ doesn't depend on the connection you started with. The second part is what I'm having trouble with.
What Morita does is suppose we are given connections $nabla^0$ and $nabla^1$ on $E$. Consider the bundle $EtimesmathbbRto MtimesmathbbR$ and define a connection $tildenabla$ on $Etimes mathbbR$ in the following way. Suppose $s$ is a section of $EtimesmathbbR$ which is independent of $t$ (that is, $s(p,t) = s(p,t')$ for all $t, t'$). Then define $tildenabla_fracpartialpartial t s = 0$ and
$$
tildenabla_X s = (1-t)nabla_X^0 s + tnabla_X^1 s
$$
for $Xin T_(p,t)(MtimesmathbbR)$.
I understand that every vector field on $Mtimes mathbbR$ can be written as a linear combination with function coefficients of $fracpartialpartial t$ and vector fields which are tangent to $Mtimes t$. Morita says that also every section of $Etimes mathbbR$ can be written as a linear combination with function coefficients of sections which are independent of $t$. I don't see why this is true.
Moreover, aren't you supposed to be feeding connections vector fields and not tangent vectors? Is the point here that a vector field on $MtimesmathbbR$ which is tangent to $Mtimes t$ can be considered as a vector field on $M$?
differential-geometry vector-bundles connections characteristic-classes
asked Jul 26 at 23:05
Ross
213
Yes, of course. I've edited that.
– Ross
Jul 26 at 23:27
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading an excerpt from Morita's book. The story is that if you have a connection $nabla$ on a vector bundle $E$ over some manifold $M$ and a function $M_n(mathbbR)to mathbbR$ which is a homogeneous polynomial invariant under conjugation, then you obtain a form on $M$ by applying $f$ to the curvature matrix $Omega$. What needs to be shown is that $f(Omega)$ is a closed form, and that $[f(Omega)]$ doesn't depend on the connection you started with. The second part is what I'm having trouble with.
What Morita does is suppose we are given connections $nabla^0$ and $nabla^1$ on $E$. Consider the bundle $EtimesmathbbRto MtimesmathbbR$ and define a connection $tildenabla$ on $Etimes mathbbR$ in the following way. Suppose $s$ is a section of $EtimesmathbbR$ which is independent of $t$ (that is, $s(p,t) = s(p,t')$ for all $t, t'$). Then define $tildenabla_fracpartialpartial t s = 0$ and
$$
tildenabla_X s = (1-t)nabla_X^0 s + tnabla_X^1 s
$$
for $Xin T_(p,t)(MtimesmathbbR)$.
I understand that every vector field on $Mtimes mathbbR$ can be written as a linear combination with function coefficients of $fracpartialpartial t$ and vector fields which are tangent to $Mtimes t$. Morita says that also every section of $Etimes mathbbR$ can be written as a linear combination with function coefficients of sections which are independent of $t$. I don't see why this is true.
Moreover, aren't you supposed to be feeding connections vector fields and not tangent vectors? Is the point here that a vector field on $MtimesmathbbR$ which is tangent to $Mtimes t$ can be considered as a vector field on $M$?
differential-geometry vector-bundles connections characteristic-classes
asked Jul 26 at 23:05
Ross
213
I am reading an excerpt from Morita's book. The story is that if you have a connection $nabla$ on a vector bundle $E$ over some manifold $M$ and a function $M_n(mathbbR)to mathbbR$ which is a homogeneous polynomial invariant under conjugation, then you obtain a form on $M$ by applying $f$ to the curvature matrix $Omega$. What needs to be shown is that $f(Omega)$ is a closed form, and that $[f(Omega)]$ doesn't depend on the connection you started with. The second part is what I'm having trouble with.
What Morita does is suppose we are given connections $nabla^0$ and $nabla^1$ on $E$. Consider the bundle $EtimesmathbbRto MtimesmathbbR$ and define a connection $tildenabla$ on $Etimes mathbbR$ in the following way. Suppose $s$ is a section of $EtimesmathbbR$ which is independent of $t$ (that is, $s(p,t) = s(p,t')$ for all $t, t'$). Then define $tildenabla_fracpartialpartial t s = 0$ and
$$
tildenabla_X s = (1-t)nabla_X^0 s + tnabla_X^1 s
$$
for $Xin T_(p,t)(MtimesmathbbR)$.
I understand that every vector field on $Mtimes mathbbR$ can be written as a linear combination with function coefficients of $fracpartialpartial t$ and vector fields which are tangent to $Mtimes t$. Morita says that also every section of $Etimes mathbbR$ can be written as a linear combination with function coefficients of sections which are independent of $t$. I don't see why this is true.
Moreover, aren't you supposed to be feeding connections vector fields and not tangent vectors? Is the point here that a vector field on $MtimesmathbbR$ which is tangent to $Mtimes t$ can be considered as a vector field on $M$?
differential-geometry vector-bundles connections characteristic-classes
asked Jul 26 at 23:05
Ross
213
edited Jul 26 at 23:27
asked Jul 26 at 23:05
Ross
213
asked Jul 26 at 23:05
Ross
213
asked Jul 26 at 23:05
Ross
213
213
Yes, of course. I've edited that.
– Ross
Jul 26 at 23:27
add a comment |Â
Yes, of course. I've edited that.
– Ross
Jul 26 at 23:27
Yes, of course. I've edited that.
– Ross
Jul 26 at 23:27
Yes, of course. I've edited that.
– Ross
Jul 26 at 23:27
add a comment |Â
1 Answer
1
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0
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Regarding your first question, it may help to note that the vector bundle $EtimesmathbbRto MtimesmathbbR$ is the pullback bundle of $Eto M$ along the projection $p:MtimesmathbbRto M$: $$beginarraycccEtimesmathbbR&to&E\downarrow&quad&downarrow\MtimesmathbbR&xrightarrowp&Mendarrayquad.$$ In particular, if $e_1,ldots,e_n$ is a local frame of $E$ on $Usubset M$, then it is a frame of $EtimesmathbbR$ on $UtimesmathbbR$, as well.
Regarding your second question, it is a fundamental fact that if the vector fields $X,X'inmathfrakX(M)$ satisfy $X(p)=X'(p)$ for some $pin M$, then for a section $s$ of $E$ we have $$nabla_Xs(p)=nabla_X's(p).$$ (This can be shown in a few different ways, depending on your definition of a connection). Hence, the expression $$nabla_vs$$ is actually well-defined for a tangent vector $v$.
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Regarding your first question, it may help to note that the vector bundle $EtimesmathbbRto MtimesmathbbR$ is the pullback bundle of $Eto M$ along the projection $p:MtimesmathbbRto M$: $$beginarraycccEtimesmathbbR&to&E\downarrow&quad&downarrow\MtimesmathbbR&xrightarrowp&Mendarrayquad.$$ In particular, if $e_1,ldots,e_n$ is a local frame of $E$ on $Usubset M$, then it is a frame of $EtimesmathbbR$ on $UtimesmathbbR$, as well.
Regarding your second question, it is a fundamental fact that if the vector fields $X,X'inmathfrakX(M)$ satisfy $X(p)=X'(p)$ for some $pin M$, then for a section $s$ of $E$ we have $$nabla_Xs(p)=nabla_X's(p).$$ (This can be shown in a few different ways, depending on your definition of a connection). Hence, the expression $$nabla_vs$$ is actually well-defined for a tangent vector $v$.
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
add a comment |Â
up vote
0
down vote
Regarding your first question, it may help to note that the vector bundle $EtimesmathbbRto MtimesmathbbR$ is the pullback bundle of $Eto M$ along the projection $p:MtimesmathbbRto M$: $$beginarraycccEtimesmathbbR&to&E\downarrow&quad&downarrow\MtimesmathbbR&xrightarrowp&Mendarrayquad.$$ In particular, if $e_1,ldots,e_n$ is a local frame of $E$ on $Usubset M$, then it is a frame of $EtimesmathbbR$ on $UtimesmathbbR$, as well.
Regarding your second question, it is a fundamental fact that if the vector fields $X,X'inmathfrakX(M)$ satisfy $X(p)=X'(p)$ for some $pin M$, then for a section $s$ of $E$ we have $$nabla_Xs(p)=nabla_X's(p).$$ (This can be shown in a few different ways, depending on your definition of a connection). Hence, the expression $$nabla_vs$$ is actually well-defined for a tangent vector $v$.
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Regarding your first question, it may help to note that the vector bundle $EtimesmathbbRto MtimesmathbbR$ is the pullback bundle of $Eto M$ along the projection $p:MtimesmathbbRto M$: $$beginarraycccEtimesmathbbR&to&E\downarrow&quad&downarrow\MtimesmathbbR&xrightarrowp&Mendarrayquad.$$ In particular, if $e_1,ldots,e_n$ is a local frame of $E$ on $Usubset M$, then it is a frame of $EtimesmathbbR$ on $UtimesmathbbR$, as well.
Regarding your second question, it is a fundamental fact that if the vector fields $X,X'inmathfrakX(M)$ satisfy $X(p)=X'(p)$ for some $pin M$, then for a section $s$ of $E$ we have $$nabla_Xs(p)=nabla_X's(p).$$ (This can be shown in a few different ways, depending on your definition of a connection). Hence, the expression $$nabla_vs$$ is actually well-defined for a tangent vector $v$.
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
Regarding your first question, it may help to note that the vector bundle $EtimesmathbbRto MtimesmathbbR$ is the pullback bundle of $Eto M$ along the projection $p:MtimesmathbbRto M$: $$beginarraycccEtimesmathbbR&to&E\downarrow&quad&downarrow\MtimesmathbbR&xrightarrowp&Mendarrayquad.$$ In particular, if $e_1,ldots,e_n$ is a local frame of $E$ on $Usubset M$, then it is a frame of $EtimesmathbbR$ on $UtimesmathbbR$, as well.
Regarding your second question, it is a fundamental fact that if the vector fields $X,X'inmathfrakX(M)$ satisfy $X(p)=X'(p)$ for some $pin M$, then for a section $s$ of $E$ we have $$nabla_Xs(p)=nabla_X's(p).$$ (This can be shown in a few different ways, depending on your definition of a connection). Hence, the expression $$nabla_vs$$ is actually well-defined for a tangent vector $v$.
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
answered Jul 27 at 8:06
Amitai Yuval
14.4k11026
14.4k11026
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
add a comment |Â
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
Thank you for your comments. Regarding your first comment, it seems like all that tells me is that sections of $E$ pull back to sections of $Etimes mathbbR$, which is already clear. I am wondering why all (global) sections can be written as a linear combination of these pulled back (global) sections. As to your second comment, thank you for pointing out that fact.
– Ross
Jul 27 at 11:49
add a comment |Â
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Yes, of course. I've edited that.
– Ross
Jul 26 at 23:27