# Inner product and projection in pseudo-riemannian manifolds

My question is about how to properly compute the projection of a tensor in a given direction using inner product product in a pseudo-riemannian manifold, since inner product isn't defined positive.

For simplicity, suppose we have a 2-dimensional curved manifold with signature $$(-, +)$$ and a generic metric tensor $$g_ {\mu \nu}$$. I want to project a time-like vector $$t ^ \mu = (t ^ 0, t ^ 1)$$ on another time-like vector $$n ^ \mu = (n ^ 0, n ^ 1)$$ and suppose to know that these two vector form an acute angle, so we expect that the projected component $$t ^ \mu_p$$ will be something like

$$t ^ \mu_p = An ^ \mu$$ with $$A> 0$$

because the projected component of $$t ^ \mu$$ has the same direction of $$n ^ \mu$$. To determine $$A$$ I suppose I have to use the inner product $$t_ \mu n ^ \mu$$, but this of course is negative, since both vectors are time-like. How do I justify mathematically the use of the minus sign that I want in the projection in order to obtain the correct result?

Then, if I want to do the same with a p-rank tensor $$T$$, do I have to consider any minus sign somewhere? For example, for a 2-rank tensor the projected component

$$T ^ {\mu \nu} _p = (T _ {\alpha \beta} n ^ \alpha n ^ \beta) n ^ \mu n ^ \nu$$

or do I need a minus sign?

The product $$n_{\mu}n^{\mu}$$ takes care of the sign without having to write it by hand, whatever the nature of $$n$$ is that is what your $$t$$ inherits. Multiply, better still, contract your first expression with $$n_{\mu}$$ then since $$n_{\mu}n^{\mu}\neq 0$$ you can get the $$A$$ straight away. Just be careful with notation:
$$n_{\mu} t_p^{\mu} = A n_{\mu}n^{\mu}\\[10pt] A = \frac{n_{\mu} t_p^{\mu}}{n_{\nu}n^{\nu}} = \frac{n_{\mu} t_p^{\mu}}{||n||^2}.$$ The sign of $$A$$ is given by this, not because you set it up front. If you force this sign you might be forcing a contradiction.