Geometric definition of the Lorentz inner product In Euclidean space one can define the dot product as projecting one vector to the other and multiply the length of the projected vector with the length of the other vector. This definition doesn't require a basis, and it is intuitive.
Is it possible to do the same in Minkowski space? So defining the inner product of two four vectors in a geometric manner without relying on a predefined basis?
 A: The Minkowski metric tensor $\mathbf \eta$ takes two four-vectors as arguments and produces a real number, the inner product of the two vectors:
$$\mathbf \eta(\vec u, \vec v) = \vec u \cdot \vec v = \langle\tilde u, \vec v\rangle$$
where the one-form $\tilde u$ is given by
$$\tilde u = \mathbf \eta (\vec u, )$$
Geometrically, this is pictured as counting the number of surfaces of the one-form $\tilde u$ 'pierced' by the vector $\vec v$.

Image credit
A: Yes and no.
Firstly projection is an operation that takes a vector and gives another vector that lives in a subspace.  Projecting doesn't give a length (though you can talk about the length of s projection).
And sometimes people talk about different kinds of projections but the first (and only) projection many people learn is the orthogonal projection, so some people think that projection is the only one and you need to watch out since people can mean different things by the word projection. But a projection always takes a vector and gives a vector.
Taking an inner product (or a bilinear form) takes two vectors and gives a scalar. One way it to talk about the length of the projection of one o to the other, but then you still need a sign.
For many situations nothing changes you can imagine the subspace you project onto and you imagine its orthogonal complement (if you projection isn't orthogonal you imagine some more general complement) then you express the vector to be projected as a sum of two vectors one from each of the two subspaces and then the projection is picking that vector from the right space.
But that fails if the thing you are projecting onto is light like. Specifically the hyperplane of vectors that are metrically orthogonal to a lightlike vector contains the light like vector.
If you tried to project a light like vector onto itself it is in the space spanned by the vector and it is also in the space of things orthogonal to the vector. It is orthogonal to itself.
So you can project is you have a complementing space, but the set if orthogonal vector does not complement the lightlike vector. So orthogonal projections fail.
Since the inner product is symmetry you could project onto the other one unless they are both lightlike. Or since the inner product is bilinear you can define it the standard way with orthogonal projections onto timelike and spacelike vectors and then use those algebraically to get the inner product on everything.
There are also other approaches where the inner product can just be intrinsic to the vectors by making a geometric product in a space of linear combinations of products of vectors that has a multiplication that is associative and linear and distributive and sends each vector to its metric square when multiplied by itself. And then the inner product appears as the symmetric product of vectors.
In that case the geometric implication falls out automatically because orthogonal vectors product out to naturally represent subspace they span and the product of a vector with a subspace gives the sum of the larger space they span (the vector part linearly independent to the space gives that) and the part spanned by the space produces the subspace of that subspace that is the orthogonal complement of that projection.
A: Yes, you can do this.  In particular if you consider a linear function which takes two vectors to the reals (or whatever field you care about), and further insist that this function is symmetric, and that its representation in any basis is invertible (so for any basis $\{\vec{e}_i\}$, the function can be represented by a matrix $g^{ij}$, and this matrix must be non-singular), then it's easy to show that such a function has most of the properties of a metric.  The property it may not have is positive / negative definiteness.
There is then a nice theorem (which is easy to prove) that you can always find a basis for the vector space such that the components of the these functions have the form $\textrm{diag}(1, \ldots, 1, -1, \ldots, -1)$.  If the signs are all the same (so all $1$s or all $-1$s) then this is the ordinary Euclidean metric.  But more interestingly there aren't very many other possible kinds of (pseudo-)metrics: you can characterise the possibilities by summing the diagonal in any basis where it has the above form, and in $4$ dimensions the only possibilities are $4$, $2$, $0$ and $-2$ (same as $2$), $-4$ (same as $4$).
The Minkowski metric is the $2$ (or $-2$) case.
Note that although I've mentioned bases here, none of this is basis dependent.
