Maximum number of vectors with pairwise negative inner product

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 $$\langle p, q \rangle = \int_0^1 p(x)q(x)dx.$$

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.

• Could you clarify? How can I map orthonormal basis of $V$ to standard basis of $R^n$? For example, orthonormal basis of space of polynomials with degree at most 2 is $e_1(x) = 1, e_2(x) = \sqrt{3}(-1 + 2x), e_3(x) = \sqrt{5}(1 - 6x + 6x^2)$. Also, I have positive-definiteness condition. I have edited my question. – edubrovskiy Nov 24 '15 at 19:43
• Ok, I got it. So, I have an isomorphism $T: V \to \mathbb{R}^n$. I can choose $n + 1$ vectors with pairwise negative inner product in $\mathbb{R}^n$ and show that for each $u, v$ of these vectors $\langle T^{-1}u, T^{-1}v \rangle < 0$. – edubrovskiy Nov 24 '15 at 20:04

The space of polynomials is infinite-dimensional. So we can choose a sequence of polynomials $p_1,p_2,p_3, \ldots$ such that each $p_{n+1}$ is perpendicular to the span of $\{p_1,p_2, \ldots, p_n\}$. That gives an increasing chain of finite-dimensional subspaces $P_n =$ span$\{p_1,p_2, \ldots, p_n\}$.

By how we chose the polynomials $p_n$ we know the inner product restricted to each $P_n$ looks like the Euclidean inner product on $\mathbb R ^n$.

Then you can choose vectors $\{v_1,v_2, v_3, \ldots \} \subset V$ such that each $\{v_1,v_2, \ldots v_k, v_{k+1}\} \subset P_k$ form the vertex of a regular simplex at the origin. It follows the set $\{v_1,v_2, \ldots v_k, v_{k+1}\}$ has mutually negative dot products. By extension the same is true for $\{v_1,v_2, v_3, \ldots \}$.

Question: What if the space is finite-dimensional, but the inner product is not positive-definite?