On the existence of the interior product on p-forms
$begingroup$
I study at a university where it is fashionable to provide a rather poor quality of teaching and claim that it is a world-leading institution. For example, lecture notes are often sloppy in their notation and sometimes expressions that haven't been used before are introduced without any explanation.
So here is a proposition from my notes:
"We just need to prove that it [the interior product] is well-defined and independent of the way we do that, which motivates the following abstract proof:
Proof We defined $wedge^p V$ as the dual space of the space of alternating p-multilinear forms on V. If M is a (p-1)-multilinear form on $V$ and $epsilon$ a linear form on V then
$$(epsilon M)(v_1, ..., v_p) = epsilon(v_1) M(v_2, ..., v_p) - epsilon(v_2) M(v_1, v_3, ..., v_p) + ...$$
is an alternating p-multilinear form. So if $alpha in wedge^p V$ we can define $i_{epsilon} alpha in wedge^{(p-1)}V$ by
$$ (i_epsilon alpha)(M) = alpha(epsilon M)$$
Take $V= T^{*}, epsilon = v in V^{*} = (T^{*})^{*} = TX$, gives the interior product. The first equation above gives the formula for working out interior products. $blacksquare$
I have three questions, mostly about interpretation:
(1) Given a smooth manifold N, we defined TN = ${(x,v): x in N, v in T_x N}$ and analogously for $T^*X$ with $T_x^*N$ in place of $T_xN$. We have shown that these are in fact vector bundles of dimension $2 dim N$. So I suspect when my lecturer wrote $V=T^*$ what he meant was $V=T^* N$ for some given manifold $N$. Is this correct?
(2) Why is $(T^*)^* = T$? If $V = T^*N$ (viewed either as a manifold or vector bundle) then how can you take the dual of this cotangent bundle?
(3) This proof implicitly relies on the assumption that vector fields $epsilon in C^infty(TN)$ are linear forms on $T^{*}N$. We defined vector fields as sections (ie smooth right inverses) of the projection $Pi: TX rightarrow X$. Why can these be viewed as linear forms on $V = T^*N$.
calculus geometry manifolds
asked Dec 18 '18 at 9:55
gengen
4722521
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add a comment |
$begingroup$
I study at a university where it is fashionable to provide a rather poor quality of teaching and claim that it is a world-leading institution. For example, lecture notes are often sloppy in their notation and sometimes expressions that haven't been used before are introduced without any explanation.
So here is a proposition from my notes:
"We just need to prove that it [the interior product] is well-defined and independent of the way we do that, which motivates the following abstract proof:
Proof We defined $wedge^p V$ as the dual space of the space of alternating p-multilinear forms on V. If M is a (p-1)-multilinear form on $V$ and $epsilon$ a linear form on V then
$$(epsilon M)(v_1, ..., v_p) = epsilon(v_1) M(v_2, ..., v_p) - epsilon(v_2) M(v_1, v_3, ..., v_p) + ...$$
is an alternating p-multilinear form. So if $alpha in wedge^p V$ we can define $i_{epsilon} alpha in wedge^{(p-1)}V$ by
$$ (i_epsilon alpha)(M) = alpha(epsilon M)$$
Take $V= T^{*}, epsilon = v in V^{*} = (T^{*})^{*} = TX$, gives the interior product. The first equation above gives the formula for working out interior products. $blacksquare$
I have three questions, mostly about interpretation:
(1) Given a smooth manifold N, we defined TN = ${(x,v): x in N, v in T_x N}$ and analogously for $T^*X$ with $T_x^*N$ in place of $T_xN$. We have shown that these are in fact vector bundles of dimension $2 dim N$. So I suspect when my lecturer wrote $V=T^*$ what he meant was $V=T^* N$ for some given manifold $N$. Is this correct?
(2) Why is $(T^*)^* = T$? If $V = T^*N$ (viewed either as a manifold or vector bundle) then how can you take the dual of this cotangent bundle?
(3) This proof implicitly relies on the assumption that vector fields $epsilon in C^infty(TN)$ are linear forms on $T^{*}N$. We defined vector fields as sections (ie smooth right inverses) of the projection $Pi: TX rightarrow X$. Why can these be viewed as linear forms on $V = T^*N$.
calculus geometry manifolds
asked Dec 18 '18 at 9:55
gengen
4722521
$endgroup$
add a comment |
$begingroup$
I study at a university where it is fashionable to provide a rather poor quality of teaching and claim that it is a world-leading institution. For example, lecture notes are often sloppy in their notation and sometimes expressions that haven't been used before are introduced without any explanation.
So here is a proposition from my notes:
"We just need to prove that it [the interior product] is well-defined and independent of the way we do that, which motivates the following abstract proof:
Proof We defined $wedge^p V$ as the dual space of the space of alternating p-multilinear forms on V. If M is a (p-1)-multilinear form on $V$ and $epsilon$ a linear form on V then
$$(epsilon M)(v_1, ..., v_p) = epsilon(v_1) M(v_2, ..., v_p) - epsilon(v_2) M(v_1, v_3, ..., v_p) + ...$$
is an alternating p-multilinear form. So if $alpha in wedge^p V$ we can define $i_{epsilon} alpha in wedge^{(p-1)}V$ by
$$ (i_epsilon alpha)(M) = alpha(epsilon M)$$
Take $V= T^{*}, epsilon = v in V^{*} = (T^{*})^{*} = TX$, gives the interior product. The first equation above gives the formula for working out interior products. $blacksquare$
I have three questions, mostly about interpretation:
(1) Given a smooth manifold N, we defined TN = ${(x,v): x in N, v in T_x N}$ and analogously for $T^*X$ with $T_x^*N$ in place of $T_xN$. We have shown that these are in fact vector bundles of dimension $2 dim N$. So I suspect when my lecturer wrote $V=T^*$ what he meant was $V=T^* N$ for some given manifold $N$. Is this correct?
(2) Why is $(T^*)^* = T$? If $V = T^*N$ (viewed either as a manifold or vector bundle) then how can you take the dual of this cotangent bundle?
(3) This proof implicitly relies on the assumption that vector fields $epsilon in C^infty(TN)$ are linear forms on $T^{*}N$. We defined vector fields as sections (ie smooth right inverses) of the projection $Pi: TX rightarrow X$. Why can these be viewed as linear forms on $V = T^*N$.
calculus geometry manifolds
asked Dec 18 '18 at 9:55
gengen
4722521
$endgroup$
I study at a university where it is fashionable to provide a rather poor quality of teaching and claim that it is a world-leading institution. For example, lecture notes are often sloppy in their notation and sometimes expressions that haven't been used before are introduced without any explanation.
So here is a proposition from my notes:
"We just need to prove that it [the interior product] is well-defined and independent of the way we do that, which motivates the following abstract proof:
Proof We defined $wedge^p V$ as the dual space of the space of alternating p-multilinear forms on V. If M is a (p-1)-multilinear form on $V$ and $epsilon$ a linear form on V then
$$(epsilon M)(v_1, ..., v_p) = epsilon(v_1) M(v_2, ..., v_p) - epsilon(v_2) M(v_1, v_3, ..., v_p) + ...$$
is an alternating p-multilinear form. So if $alpha in wedge^p V$ we can define $i_{epsilon} alpha in wedge^{(p-1)}V$ by
$$ (i_epsilon alpha)(M) = alpha(epsilon M)$$
Take $V= T^{*}, epsilon = v in V^{*} = (T^{*})^{*} = TX$, gives the interior product. The first equation above gives the formula for working out interior products. $blacksquare$
I have three questions, mostly about interpretation:
(1) Given a smooth manifold N, we defined TN = ${(x,v): x in N, v in T_x N}$ and analogously for $T^*X$ with $T_x^*N$ in place of $T_xN$. We have shown that these are in fact vector bundles of dimension $2 dim N$. So I suspect when my lecturer wrote $V=T^*$ what he meant was $V=T^* N$ for some given manifold $N$. Is this correct?
(2) Why is $(T^*)^* = T$? If $V = T^*N$ (viewed either as a manifold or vector bundle) then how can you take the dual of this cotangent bundle?
(3) This proof implicitly relies on the assumption that vector fields $epsilon in C^infty(TN)$ are linear forms on $T^{*}N$. We defined vector fields as sections (ie smooth right inverses) of the projection $Pi: TX rightarrow X$. Why can these be viewed as linear forms on $V = T^*N$.
calculus geometry manifolds
calculus geometry manifolds
asked Dec 18 '18 at 9:55
gengen
4722521
asked Dec 18 '18 at 9:55
gengen
4722521
asked Dec 18 '18 at 9:55
gengen
4722521
asked Dec 18 '18 at 9:55
gengen
4722521
asked Dec 18 '18 at 9:55
gengen
4722521
4722521
add a comment |
add a comment |
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