Identifying a wedge-to-metric formula
$begingroup$
In this question, the original poster wrote:
On every Riemannian manifold $M$, we can consider the Hodge
$*$-operator, which is characterised by the following formula:
$$awedge *b = (a,b)nu.$$ Here $a$ and $b$ are smooth forms on $M$,
$( , )$ is a metric on $wedge T^*!M$ and $nu$ is the volume form
with respect to the Riemannian metric.
I'm looking to study this formula in particular, but it's difficult to search for because of the notation.
What are a couple webpages or books that discuss (or even derive) this formula? The simpler the better.
differential-geometry reference-request exterior-algebra
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
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add a comment |
$begingroup$
In this question, the original poster wrote:
On every Riemannian manifold $M$, we can consider the Hodge
$*$-operator, which is characterised by the following formula:
$$awedge *b = (a,b)nu.$$ Here $a$ and $b$ are smooth forms on $M$,
$( , )$ is a metric on $wedge T^*!M$ and $nu$ is the volume form
with respect to the Riemannian metric.
I'm looking to study this formula in particular, but it's difficult to search for because of the notation.
What are a couple webpages or books that discuss (or even derive) this formula? The simpler the better.
differential-geometry reference-request exterior-algebra
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
$endgroup$
$begingroup$
Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $mathbb{R}^3$ of $U times V^* = sin alpha |U||V|$ where $V^*$ is the vector directly orthogonal to V in TM and $alpha$ the angle between $U$ and $V^*$.
$endgroup$
– Jean Marie
Dec 5 '18 at 6:53
$begingroup$
@JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example?
$endgroup$
– Doubt
Dec 5 '18 at 14:22
$begingroup$
I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form.
$endgroup$
– Jean Marie
Dec 5 '18 at 15:42
add a comment |
$begingroup$
In this question, the original poster wrote:
On every Riemannian manifold $M$, we can consider the Hodge
$*$-operator, which is characterised by the following formula:
$$awedge *b = (a,b)nu.$$ Here $a$ and $b$ are smooth forms on $M$,
$( , )$ is a metric on $wedge T^*!M$ and $nu$ is the volume form
with respect to the Riemannian metric.
I'm looking to study this formula in particular, but it's difficult to search for because of the notation.
What are a couple webpages or books that discuss (or even derive) this formula? The simpler the better.
differential-geometry reference-request exterior-algebra
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
$endgroup$
In this question, the original poster wrote:
On every Riemannian manifold $M$, we can consider the Hodge
$*$-operator, which is characterised by the following formula:
$$awedge *b = (a,b)nu.$$ Here $a$ and $b$ are smooth forms on $M$,
$( , )$ is a metric on $wedge T^*!M$ and $nu$ is the volume form
with respect to the Riemannian metric.
I'm looking to study this formula in particular, but it's difficult to search for because of the notation.
What are a couple webpages or books that discuss (or even derive) this formula? The simpler the better.
differential-geometry reference-request exterior-algebra
differential-geometry reference-request exterior-algebra
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
asked Dec 5 '18 at 4:23
DoubtDoubt
753321
753321
$begingroup$
Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $mathbb{R}^3$ of $U times V^* = sin alpha |U||V|$ where $V^*$ is the vector directly orthogonal to V in TM and $alpha$ the angle between $U$ and $V^*$.
$endgroup$
– Jean Marie
Dec 5 '18 at 6:53
$begingroup$
@JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example?
$endgroup$
– Doubt
Dec 5 '18 at 14:22
$begingroup$
I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form.
$endgroup$
– Jean Marie
Dec 5 '18 at 15:42
add a comment |
$begingroup$
Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $mathbb{R}^3$ of $U times V^* = sin alpha |U||V|$ where $V^*$ is the vector directly orthogonal to V in TM and $alpha$ the angle between $U$ and $V^*$.
$endgroup$
– Jean Marie
Dec 5 '18 at 6:53
$begingroup$
@JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example?
$endgroup$
– Doubt
Dec 5 '18 at 14:22
$begingroup$
I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form.
$endgroup$
– Jean Marie
Dec 5 '18 at 15:42
$begingroup$
Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $mathbb{R}^3$ of $U times V^* = sin alpha |U||V|$ where $V^*$ is the vector directly orthogonal to V in TM and $alpha$ the angle between $U$ and $V^*$.
$endgroup$
– Jean Marie
Dec 5 '18 at 6:53
$begingroup$
Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $mathbb{R}^3$ of $U times V^* = sin alpha |U||V|$ where $V^*$ is the vector directly orthogonal to V in TM and $alpha$ the angle between $U$ and $V^*$.
$endgroup$
– Jean Marie
Dec 5 '18 at 6:53
$begingroup$
@JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example?
$endgroup$
– Doubt
Dec 5 '18 at 14:22
$begingroup$
@JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example?
$endgroup$
– Doubt
Dec 5 '18 at 14:22
$begingroup$
I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form.
$endgroup$
– Jean Marie
Dec 5 '18 at 15:42
$begingroup$
I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form.
$endgroup$
– Jean Marie
Dec 5 '18 at 15:42
add a comment |
1 Answer
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This formula appears frequently as the definition of the Hodge star operator.
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
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add a comment |
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$begingroup$
This formula appears frequently as the definition of the Hodge star operator.
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
$endgroup$
add a comment |
$begingroup$
This formula appears frequently as the definition of the Hodge star operator.
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
$endgroup$
add a comment |
$begingroup$
This formula appears frequently as the definition of the Hodge star operator.
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
$endgroup$
This formula appears frequently as the definition of the Hodge star operator.
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
answered Dec 5 '18 at 14:35
DoubtDoubt
753321
753321
add a comment |
add a comment |
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$begingroup$
Even if this is not a direct answer to your question : intuitively, this formula can be seen as a generalization in the context of an embedded surface in $mathbb{R}^3$ of $U times V^* = sin alpha |U||V|$ where $V^*$ is the vector directly orthogonal to V in TM and $alpha$ the angle between $U$ and $V^*$.
$endgroup$
– Jean Marie
Dec 5 '18 at 6:53
$begingroup$
@JeanMarie Would you say that this formula only applies in cases where $a$ and $b$ are forms of the same degree, like in your example?
$endgroup$
– Doubt
Dec 5 '18 at 14:22
$begingroup$
I would say : yes because (.,.) is bound to be a bilinear or sesquilinear form.
$endgroup$
– Jean Marie
Dec 5 '18 at 15:42