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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.

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  • $\begingroup$ Welcome to Math.SE! In order to get higher-quality answers to this question, I'd recommend clarifying what you mean by an "indefinite inner product". This will help ensure everyone understands one another and that you get answers that suit your particular needs. Are you referring to an indefinite inner product as defined on the following Wikipedia page? en.wikipedia.org/wiki/Indefinite_inner_product_space If so, I can provide an answer later. (In fact, one can glean an answer from the first section of that article!) $\endgroup$
    – Dan
    Dec 21, 2016 at 20:21
  • $\begingroup$ Hi, thanks. By an indefinite inner product i do mean an inner product defined for the Hilbert space which statisfies all the conditions for a standard inner product except that V=0 implies [V,V]=0 but the converse is not necessarily true. $\endgroup$
    – sanaz mat
    Dec 21, 2016 at 20:51

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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.

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  • $\begingroup$ Thanks. After reading your answer i realized that the actual type of inner product which i was concerned about in my question is the positive semidefinite inner product. $\endgroup$
    – sanaz mat
    Dec 21, 2016 at 23:23
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    $\begingroup$ Happy to help! : ) Sometimes the hard part (in functional analysis and beyond) is figuring out the right level of generality is to ask a question. Glad you found it! And yes, for semi definite inner products you can see that the problems I discussed all go away. I'm on mobile but believe the link I provided has a full proof in that case. $\endgroup$
    – Dan
    Dec 21, 2016 at 23:23

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