I Hodge dual

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<Moderator's note: Moved from another forum.>

The book I'm reading says that ##\star \sigma = 1## and ##\star 1 = \sigma##, but I'm not sure about the last one. The space is ##V = \mathbb{M}^4## and we choose the canonical base ##e_0,e_1,e_2,e_3##. This means that ##g_{ij} = \mathrm{diag}(-1,1,1,1)##, so its determinant is ##-1##. We also choose ##\sigma=e_0 \wedge e_1 \wedge e_2 \wedge e_3##.
In general, given the definitions above: $$
\begin{align*}
\eta \wedge \star\lambda &= -g(\eta,\lambda)\sigma \\
\star\star\lambda &= (-1)^{p(4-p)+1}\lambda,\qquad\lambda \in \bigwedge\nolimits^p V
\end{align*}
$$ If I apply either of these equations to ##\star 1##, I get ##-\sigma##. For instance: $$
1 \wedge \star 1 = -g(1,1)\sigma = -\sigma
$$ I get the same thing even by using the definition: $$
\begin{align*}
1 \wedge \sigma &= g(\star 1, \sigma)\sigma &\implies \\
g(\star 1, \sigma) &= 1 &\implies \\
\star 1 &= -\sigma
\end{align*}
$$
 
Last edited by a moderator:

Cryo

Gold Member
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It is more of a question then answer - sorry. How would you compute

##1 \wedge \sigma## ?

It sort of makes sense, but I do not know how to tackle it. The reason I have this question is that to solve your first equation ##\eta \wedge \star \lambda \dots##, for ##\lambda=\sigma##, I would need to consider this. Could ##\star 1=\sigma## be a convention?
 
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It is more of a question then answer - sorry. How would you compute

##1 \wedge \sigma## ?

It sort of makes sense, but I do not know how to tackle it. The reason I have this question is that to solve your first equation ##\eta \wedge \star \lambda \dots##, for ##\lambda=\sigma##, I would need to consider this. Could ##\star 1=\sigma## be a convention?
I assumed ##1\wedge\sigma=\sigma## but it doesn't make much sense. Maybe ##\star 1=\sigma## is just a convention.
 
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Hodge dual for scalars (0 forms) is just multiplication, so it does make perfect sense for ##1 \wedge \omega = \omega## (That is, ## f \wedge a = fa## if f is a 0-form, and a is any form).

I have no clue why you're making your inner product negative. It all falls into place if you say ## a \wedge \star b = g(a,b) \sigma##

Now, let's ask what ##\star 1## is. We know that ##a \wedge \star a = g(a,a) \sigma##. Thus, ##1 \wedge \star 1 = g(1,1) \sigma = \sigma## Since 1 is a 0-form, the only conclusion we can make is ##\star 1 = \sigma##

I have some posts in my history computing other forms, but i'm not a mathematician so there isnt much rigor.

Also, i think what might be tripping you up is the ##\star \sigma = 1## because on lorentzian manifolds, this isn't the case because the bigger picture is ## \sigma \wedge \star \sigma = g(\sigma, \sigma) \sigma = (-1)^s \sigma## where s is the signature of your metric, and in your case, that is 1, so ##\star \sigma = -1## on those manifolds.

EDIT: To be even more clear, ##\star \sigma = (-1)^s##.
 
Last edited by a moderator:
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I'm making the inner product negative because if ##g(\sigma,\sigma)=(-1)^d##, then ##a\wedge\star b = (-1)^d g(a,b)\sigma##, according to the book. Here's the full theorem:

Let ##V## be ##n##-dimensional with inner product ##g##. Let ##\eta,\lambda \in \bigwedge\nolimits^p V##, and choose ##\sigma \in \bigwedge\nolimits^n V## to satisfy ##g(\sigma, \sigma)=(-1)^d##. Then $$
\star\star\lambda = (-1)^{p(n-p)+d} \lambda,
$$ and $$
\eta \wedge \star\lambda = \lambda \wedge \star\eta = (-1)^d g(\eta,\lambda)\sigma.
$$

In general, the Hodge dual is defined through this relation: $$
\lambda\wedge\mu = g(\star\lambda, \mu)\sigma.
$$ If ##g_{ij} = \mathrm{diag}(-1,1,1,1)##, we must have ##g(\sigma,\sigma)=\det(g_{ij})=-1## as the book states. If it makes sense to say that ##g(1,1)=1## and the definitions keep working, then $$
\sigma\wedge 1 = g(\star\sigma, 1)\sigma \implies g(\star\sigma,1)=1 \implies \star\sigma = 1
$$ Similarly, $$
1 \wedge \sigma = g(\star1, \sigma)\sigma \implies g(\star 1, \sigma)=1 \implies \star 1 = -\sigma
$$ because ##g(\sigma,\sigma)=-1##.
Do you see anything wrong with my reasoning? Are we using the same ##\sigma##? In what I wrote, ##\sigma=e_0\wedge e_1\wedge e_2\wedge e_3##, where ##e_0=(1,0,0,0), e_1=(0,1,0,0), e_2=(0,0,1,0)##, and ##e_3=(0,0,0,1)##. The Lorentzian inner product was defined as $$
g(u,v) := -u_0 v_0 + \sum_{i=1}^{n-1} u_i v_i.
$$ which means that ##g(e_0,e_0)=-1## and thus ##g(\sigma,\sigma)=-1##.
 
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I'm not sure what book you're learning this from, but I have never seen it being ##g(\star a, a)##, I've only seen ##g(\star a, \star a)## or ##g(a,a)##

Regardless, your logic should be fine since you're consistent. That is, ##\star 1 = -\sigma## and ##\star \sigma = 1## whereas I get ##\star 1 = \sigma## and ##\star \sigma = -1## so we're off by a factor of ##(-1)^s## which is fine.

The book i learned the basics of this from has it online for free, so you can check it out here, this page should talk more about it: http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/hodge

Maybe this one will be closer to what you want: http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/hodge2
 
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