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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*}

$$

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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