Signature of the inner product. I need some help with this problem. Consider in $\mathbb{R}^{4}$ the lorentz's metric $h$ which has a signature $(3,1)$, this mean that diagonal matrix $M(h)$ is the form $$M(h)=\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1\\
\end{pmatrix}$$
How I can find the signature of the inner product, induce by $h$ in $\bigwedge ^{3}(\mathbb{R}^{4})$??
In my textbook, say for example, if $h$ is a scalar product with signature $(n-1,1)$ in $V$, then the induced scalar product in the space $\bigwedge^{k}(V^{*})$ of the exterior k-forms has signature $({n-1\choose k},{n-1\choose k-1})$, but I not understand this, any hint. Thanks!
 A: One way to prove the signature of the induced metric is to use clifford algebra: an algebra that incorporates both the structure of the exterior algebra as well as that of the inner product.
The clifford product, denoted by juxtaposition (that is, $ab$ is the clifford product of $a$ and $b$) obeys the following relations: for one-forms $\alpha, \beta, \gamma$,
$$\alpha \beta + \beta \alpha = 2 \langle \alpha, \beta \rangle , \quad (\alpha \beta) \gamma = \alpha (\beta \gamma)$$
Note that the first relation implies $\alpha \alpha = \langle \alpha, \alpha \rangle$. In addition, when $\alpha \perp \beta$, $\alpha \beta = -\beta \alpha$, like in the exterior algebra.
As with the exterior algebra, elements ("multivectors") of the clifford algebra can be partitioned into grades.  Any multivector that can be written as a linear combination of products of $k$ orthogonal vectors (or one-forms) is a "grade-$k$ multivector". Moreover, any general multivector can be written as a linear combination of singly graded multivectors: in this case, as a combination of grade-0, -1, -2, -3, and -4 multivectors.
Further, we can identify the set of grade-3 multivectors on $\mathbb R^{3,1}$--call this $\mathcal{Cl}^3 {\mathbb R^{3,1}}$--with $\bigwedge^3 R^{3,1}$.

With that in mind, now we can directly prove the signature of the induced metric on 3-forms. We can do this using the associativity of the clifford product.  Let $\theta^1, \theta^2, \theta^3, \theta^4$ be the standard basis for 1-forms.  Then any 3-form can be written as a linear combination of
$$\theta^1 \theta^2 \theta^3, \theta^1 \theta^2 \theta^4, \theta^2 \theta^3 \theta^4, \theta^1 \theta^3 \theta^4$$
Again, this is using the clifford product, but because the standard basis is orthogonal, these products here are indistinguishable from the exterior product.
How do we use this to get the induced inner product?
Assertion: the induced inner product of two multivectors $A, B$ is the grade-0 part of the product $A B^\dagger$, denoted $[AB^\dagger]_0$, where $B^\dagger$ is the "reverse" of $B$.
Therefore, we find that the inner product $\langle \theta^1 \theta^2 \theta^3, \theta^1 \theta^2 \theta^3\rangle$ is
$$\langle \theta^1 \theta^2 \theta^3, \theta^1 \theta^2 \theta^3\rangle = [\theta^1 \theta^2 \theta^3 \theta^3 \theta^2 \theta^1]_0$$
where the brackets $[X]_0$ mean "take the grade-0 part of the multivector $X$.
Using associativity, and the relation $\alpha \alpha = \langle \alpha, \alpha \rangle$ for any vector/1-form, as well as the anticommutativity of orthogonal vectors/forms, the result is $+1$.
On the other hand, some of these products produce no scalar part: consider
$$\langle \theta^1 \theta^2 \theta^3, \theta^2 \theta^3 \theta^4 \rangle = [\theta^1 \theta^2 \theta^3 \theta^4 \theta^3 \theta^2 ]_0 = [\theta^1 \theta^4]_0 = 0$$
So using this method, we can brute force show that the metric signature must be $(1, 3)$ as given.

In the general case, with any number of Euclidean dimensions, and considering any particular grade $k$, here is the logic to use:
In $n$ dimensions, there are $\binom{n}{k}$ basis forms.  Of these, $\binom{n-1}{k}$ are totally Euclidean forms--none of them involve the basis form with the opposite sign in the metric. All of these have the same signature. Then, the only other basis forms left are those that have the timelike basis form appearing exactly once in the product. The remaining $k-1$ basis forms are chosen from the spacelike basis one-forms, hence $\binom{n-1}{k-1}$ ways to choose these.
All that remains is to figure out what the resulting signature is. Using the assertion I made above, the clifford algebra relation consistent with your convention will always yield $+1$ for forms composed of only spacelike basis forms, and always $-1$ for forms with the single timelike basis form.
