Is a metric's form determined by its signature? Suppose that we define a 4-dimensional vector space over the real field with a metric with signature (3, 1). Is the scalar product map determined only with this information? 
For example: a Minowsky space is defined as a space like this with a scalar product defined as
$$ a \cdot b = a_0b_0 - a_1b_1 - a_2b_2 - a_3b_3$$ 
My question is: could we leave the scalar product out of the definition? Would the (3, 1) metric imply that the scalar product has this form? 
 A: Let's write the definition in matrix form:
$$ a\cdot b = (a_0\;a_1\;a_2\;a_3)\begin{pmatrix}1\\&-1\\&&-1\\&&&-1\end{pmatrix} \begin{pmatrix}b_0\\b_1\\b_2\\b_3\end{pmatrix} $$
To say that some inner product $\langle a,b\rangle$ has signature $(3,1)$ is to say that the inner product looks like this dot product up to linear transformations -- that is, that there is some invertible matrix $P$ such that
$$ \langle a,b\rangle = (a_0\;a_1\;a_2\;a_3)P^T \begin{pmatrix}1\\&-1\\&&-1\\&&&-1\end{pmatrix} P \begin{pmatrix}b_0\\b_1\\b_2\\b_3\end{pmatrix} $$
(Sylvester's Law says that the signature of a symmetric bilinear form according to this definition is unique).
The Lorentz transformations are exactly those that leave the Minkowski dot product unchanged -- that is, $P$ the matrix of a Lorentz transformation if and only if
$$ P^T \begin{pmatrix}1\\&-1\\&&-1\\&&&-1\end{pmatrix} P = \begin{pmatrix}1\\&-1\\&&-1\\&&&-1\end{pmatrix} $$
[Up to minor conventions: some authors don't consider a transformation to be Lorentz if it includes a time reversal and/or a space parity inversion. And some would say $(1,3)$ instead and/or have the time coordinate be the last elements in the 4-vectors, and/or invert the signs of the products.]
