# I Hodge dual

#### kiuhnm

<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*}

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#### Cryo

Gold Member
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?

#### kiuhnm

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.

#### romsofia

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, $\star \sigma = (-1)^s$.

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#### kiuhnm

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 $g(\sigma,\sigma)=-1$.

#### romsofia

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

#### kiuhnm

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 $g(a,a)$
I'm reading Renteln's book.