Does an indefinite inner product induce a seminorm? I know that an inner product defined for a vector space induces a norm and the norm induces a metric and the metric will eventually induce a topology. I'm also aware that a seminorm induces  a pseudometric and that again induces a pseudometric topology. Now I'm curious weather an indefinite inner product induces a seminorm or not?
By an indefinite inner product I do mean an inner product defined for the Hilbert space which satisfies all the conditions for a standard inner product except that V=0 implies [V,V]=0 but the converse is not necessarily true.
 A: Thanks for providing a definition. I'll go ahead and assume that we're working with vector spaces defined over $\mathbb{C}$, but most of what I say applies equally well over subfields over other archimedean valued fields (in particular, over subfields of $\mathbb{C}$). The short answer is: the usual definition $p(v) = \langle v,v\rangle^{1/2}$ does not produce a seminorm, but modifications such as $p(v) = |\langle v,v\rangle|^{1/2}$ do.
More precisely, let $(V,\langle\cdot,\cdot\rangle)$ denote an indefinite inner product space, in the sense you gave in the comments and for each $v \in V$ define $p(v) = \langle v,v\rangle^{1/2}$. Then it is shown, for instance, here, that $p$ is absolutely homogeneous ($p(\alpha v) = |\alpha|p(v)$ for all $\alpha \in \mathbb{C}$ and $v \in V$) using the usual follow-your-nose proof from the definition of $\langle\cdot,\cdot\rangle$. The "classical" proofs of the other properties required for $p$ to be a seminorm depend on $\langle v,v\rangle \in [0,\infty)$ for all $v \in \mathbb{R}$, which is not the case in general for an indefinite inner product. In fact, the even bigger problem is that this definition of $p$ might not even be well-defined, since $\langle v,v\rangle$ can be negative.
All of these problems do, however, go away if we set $p(v) = |\langle v,v\rangle|^{1/2}$ instead, and the usual proofs of positive semi-definiteness and subadditivity apply, proving that this modified $p$ is a seminorm.
More on indefinite inner products (and links to textbooks covering them) can be found in this older Math.SE thread.
