Let $V$ be a vector space with positive-definite inner product
$$\langle \cdot, \cdot \rangle: V \times V \to \mathbb{R}.$$
Let $\dim V = n$, $n \in \mathbb{Z}$. What is the largest number $m$, such that there exist $v_1, \dots, v_m \in V$ such that $\langle v_i, v_j\rangle < 0$ for all $i \neq j$?
I know that answer is $n + 1$ if $V$ is $\mathbb{R}^n$ with standard inner product. Here is the proof:
https://mathoverflow.net/questions/31436/largest-number-of-vectors-with-pairwise-negative-dot-product
What if $V$ is arbitrary? For example, let $V$ be a space of polynomials with degree at most $k$ and define inner product by
$$\langle p, q \rangle = \int_0^1 p(x)q(x)dx.$$
Best Answer
If you have a positive-definiteness condition on your inner product, then the answer is the same for any $n$-dimensional real vector space (with $n$ finite). You can see this by choosing an orthonormal basis for $V$ and mapping it to the standard basis in $\Bbb R^n$, extending this map linearly; the two are then isomorphic as inner product spaces.
If you relax positive-definiteness then the answer depends on the signature of the inner product, and can be infinite. For example, in Minkowski space you can have vectors such that $\langle v,v\rangle <0$. For such a $v$, every element of the set $\{\lambda v\}$ with $\lambda>0$ has negative inner product with every other element of the set.