# Vector spaces without an inner product

Many simple vector spaces such as $$\mathbb{R}^m$$ have natural inner products, in this case the dot product. Even function spaces such as $$C[0,1]$$, continuous real-valued functions on the unit interval, lend themselves to nice inner products, e.g. $$\langle f,g \rangle := \int^1_0 f(x)g(x) dx.$$

In trying to think of vector spaces $$V$$ without a sensible inner product, it seems easiest to put $$V$$ over a less "nice" field, say the finite field $$\mathbb{F}_5$$. If $$V$$ is the vector space of polynomials up to degree $$n$$ with coefficients in $$\mathbb{F}_5$$, what would a reasonable inner product on $$V$$ be? What's an example of a real vector space without an inner product?

More generally, given an $$F$$-vector space $$V$$, are there necessary and sufficient conditions to show $$V$$ has an inner product?

• an inner product should be positive-definite, so in an ordered field Jun 7, 2020 at 21:42

If you don't assume choice, then this answer show that it is consistent with $$\sf ZF+ DC$$ that $$\cal C(\Bbb R)$$ has no norm. (Since every inner product gives a norm, it answers your question as well.)
Note that inner-product requires your base field to have an "order", that is, your field must be an ordered field. In particular, it must have characteristic $$0$$. Therefore, it doesn't make much sense to talk about finite fields.
• Thanks! I suspected we needed $F$ to have characteristic $0$, since there is a natural connection between norms and inner products and so we need the triangle inequality to hold. Jun 8, 2020 at 23:43