Hodge star operation, Laplacian and $L_2$ norm in relation to orientability of manifold
$begingroup$
Let $M$ be a $n$ dimensional smooth manifold.(Don't know whether $M$ is orientable or not.)
Define $a,binOmega^k(M)$. Then adopt standard Hodge star definition. $<a,b>=star(awedgestar b)$ and one can integrate $awedgestar b$ over $M$ to define $L_2$ norm. Similarly one can define adjoint $d^star$ to $d$ both of which are independent of orientabilility. Then $Delta=dd^star+d^star d$ which is also independent of orientation.
$textbf{Q:}$ How come $L_2$ "norm" is a norm in non-orientable case here? Since I do not have notion of good volume element, it is possible $int_M dV=0$ for some "$dV$" top form. What if this top form comes from $awedgestar a$ for some $ainOmega^k(M)$? Shouldn't this violate non-degeneracy of norm?
The following is the quote of the book. "Since two stars appear in $d^star$ and hence also $Delta$ may be defined for non-orientable Riemann manifolds. We just define it locally, hence globally up to a choice of sign which then cancels in $d^star=(-1)^{?}star dstar$. Similarly $L_2$ product can be defined on non-orientable Riemann manifold, because of ambiguity of sign of $star$ involved cancels the one coming from integration."
$textbf{Q':}$ How come global choice of sign is cancelled out in $d^star=(-1)^{?}star dstar$ and $L_2$ product is well defined? Furthermore why is ambiguity of sign of $star$ involved cancels the one coming from integration?
Ref. Jost Riemannian Geometry and Geometric Analysis Chpt 3's Rmk right after Definition 3.3.2
geometry differential-geometry
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
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add a comment |
$begingroup$
Let $M$ be a $n$ dimensional smooth manifold.(Don't know whether $M$ is orientable or not.)
Define $a,binOmega^k(M)$. Then adopt standard Hodge star definition. $<a,b>=star(awedgestar b)$ and one can integrate $awedgestar b$ over $M$ to define $L_2$ norm. Similarly one can define adjoint $d^star$ to $d$ both of which are independent of orientabilility. Then $Delta=dd^star+d^star d$ which is also independent of orientation.
$textbf{Q:}$ How come $L_2$ "norm" is a norm in non-orientable case here? Since I do not have notion of good volume element, it is possible $int_M dV=0$ for some "$dV$" top form. What if this top form comes from $awedgestar a$ for some $ainOmega^k(M)$? Shouldn't this violate non-degeneracy of norm?
The following is the quote of the book. "Since two stars appear in $d^star$ and hence also $Delta$ may be defined for non-orientable Riemann manifolds. We just define it locally, hence globally up to a choice of sign which then cancels in $d^star=(-1)^{?}star dstar$. Similarly $L_2$ product can be defined on non-orientable Riemann manifold, because of ambiguity of sign of $star$ involved cancels the one coming from integration."
$textbf{Q':}$ How come global choice of sign is cancelled out in $d^star=(-1)^{?}star dstar$ and $L_2$ product is well defined? Furthermore why is ambiguity of sign of $star$ involved cancels the one coming from integration?
Ref. Jost Riemannian Geometry and Geometric Analysis Chpt 3's Rmk right after Definition 3.3.2
geometry differential-geometry
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $n$ dimensional smooth manifold.(Don't know whether $M$ is orientable or not.)
Define $a,binOmega^k(M)$. Then adopt standard Hodge star definition. $<a,b>=star(awedgestar b)$ and one can integrate $awedgestar b$ over $M$ to define $L_2$ norm. Similarly one can define adjoint $d^star$ to $d$ both of which are independent of orientabilility. Then $Delta=dd^star+d^star d$ which is also independent of orientation.
$textbf{Q:}$ How come $L_2$ "norm" is a norm in non-orientable case here? Since I do not have notion of good volume element, it is possible $int_M dV=0$ for some "$dV$" top form. What if this top form comes from $awedgestar a$ for some $ainOmega^k(M)$? Shouldn't this violate non-degeneracy of norm?
The following is the quote of the book. "Since two stars appear in $d^star$ and hence also $Delta$ may be defined for non-orientable Riemann manifolds. We just define it locally, hence globally up to a choice of sign which then cancels in $d^star=(-1)^{?}star dstar$. Similarly $L_2$ product can be defined on non-orientable Riemann manifold, because of ambiguity of sign of $star$ involved cancels the one coming from integration."
$textbf{Q':}$ How come global choice of sign is cancelled out in $d^star=(-1)^{?}star dstar$ and $L_2$ product is well defined? Furthermore why is ambiguity of sign of $star$ involved cancels the one coming from integration?
Ref. Jost Riemannian Geometry and Geometric Analysis Chpt 3's Rmk right after Definition 3.3.2
geometry differential-geometry
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
$endgroup$
Let $M$ be a $n$ dimensional smooth manifold.(Don't know whether $M$ is orientable or not.)
Define $a,binOmega^k(M)$. Then adopt standard Hodge star definition. $<a,b>=star(awedgestar b)$ and one can integrate $awedgestar b$ over $M$ to define $L_2$ norm. Similarly one can define adjoint $d^star$ to $d$ both of which are independent of orientabilility. Then $Delta=dd^star+d^star d$ which is also independent of orientation.
$textbf{Q:}$ How come $L_2$ "norm" is a norm in non-orientable case here? Since I do not have notion of good volume element, it is possible $int_M dV=0$ for some "$dV$" top form. What if this top form comes from $awedgestar a$ for some $ainOmega^k(M)$? Shouldn't this violate non-degeneracy of norm?
The following is the quote of the book. "Since two stars appear in $d^star$ and hence also $Delta$ may be defined for non-orientable Riemann manifolds. We just define it locally, hence globally up to a choice of sign which then cancels in $d^star=(-1)^{?}star dstar$. Similarly $L_2$ product can be defined on non-orientable Riemann manifold, because of ambiguity of sign of $star$ involved cancels the one coming from integration."
$textbf{Q':}$ How come global choice of sign is cancelled out in $d^star=(-1)^{?}star dstar$ and $L_2$ product is well defined? Furthermore why is ambiguity of sign of $star$ involved cancels the one coming from integration?
Ref. Jost Riemannian Geometry and Geometric Analysis Chpt 3's Rmk right after Definition 3.3.2
geometry differential-geometry
geometry differential-geometry
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
asked Nov 29 '18 at 2:04
user45765user45765
2,6942722
2,6942722
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