Signature of the norm on $k$-forms. I wonder about the signature of the inner product on forms on a vector space equipped with a non-degenerate bilinear of signature $(t,s)$. For definiteness, let $(V_{6}, g)$ be an oriented six-dimensional vector space equipped with a non-degenerate metric $g$ of signature $(s,t)$ and let $W=\Lambda^{3}V^{\ast}_{6}$. In $W$ there is an induced metric $<\cdot,\cdot>$ given by
$< h_{1}, h_{2}> \Omega = h_{1}\wedge \ast h_{2}\, , \qquad h_{1}, h_{2} \in W$
where $\Omega$ is the standard volume form induced by $g$ and $\ast$ is the Hodge dual associated to $g$. I would like to know what is the signature of $<\cdot,\cdot>$ assuming that the signature of $g$ is $(t,s)$. If $t=6$ and $s=0$ then $<\cdot,\cdot>$ is definite positive, but that is not the case in general. For example, if $(t,s)=(1,5)$ then 
$< h, h> = 0$ if $h = \ast h$, namely if $h\in W$ is self-dual (by the way, is this an if and only if?)
Thanks
 A: Using the Hodge-$*$ in this case seems rather distracting to me. It is easier to observe that the induced inner product on $\Lambda^k V$ can be characterized by the fact that $\langle v_1\wedge\dots\wedge v_k,w_1\wedge\dots\wedge w_k\rangle=\det(\langle v_i,w_j\rangle)$ for decomposable elements. In particular, if you start with an orthonormal basis of $V$ (so you have $\langle v_i,v_j\rangle=\delta_{ij}\epsilon_i$ with $\epsilon_i=\pm 1$), then the induced basis for $\Lambda^k V$ will be orthonormal, too. Moreover, then of course $\langle v_{i_1}\wedge\dots\wedge v_{i_k}, v_{i_1}\wedge\dots\wedge v_{i_k}\rangle=\epsilon_{i_1}\dots\epsilon_{i_k}$. From this you can read off the signature directly. Alternatively, if you write $V=V'\oplus V''$ (orthgonal sum) such that the initial inner product is positive definite on $V'$ and negative definite on $V''$, then the induced decomposition $\Lambda^kV=\oplus_{i+j=k}(\Lambda^iV')\otimes(\Lambda^jV'')$ is orthogonal, and the induced inner product on $(\Lambda^iV')\otimes(\Lambda^jV'')$ is positive definite for even $j$ and negative definite for odd $j$. 
In the specific case you are asking ($\dim(V)=6$, $(t,s)=(1,5)$) the decomposition is $\Lambda^3 V=(V'\otimes\Lambda^2 V'')\oplus \Lambda^3V''$ with the summands having dimension 10, so the signature is $(10,10)$ in this case. If you start with $(t,s)=(2,4)$ instead, then you get $\Lambda^3 V=(\Lambda^2V'\otimes V'')\oplus (V'\otimes\Lambda^2 V'')\oplus \Lambda^3V''$ with dimensions $4$, $2\times 6=12$ and $4$, so the signature becomes $(12,8)$ in this case, and so on. 
