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I am working with linear algebra over finite fields, specifically $F_2$. In class my professor has explained that every inner product induces a norm, $\sqrt{\left < v,v \right>}$ which in turn induces a metric. All of this has seemed pretty obvious to me.

Considering the vector space $V={F_2}^2$, with the standard sum of the product of each coordinate inner product, it seems to me that this is not true (probably because I am not understanding something right). The inner product is a function from $V \times V \rightarrow V$ which means that the under the induced metric, the magnitude of $\left(1,1 \right)^T$ is zero, though it is not the zero vector, contradicting the definition of a metric.

Can somebody please clear up my misunderstanding?

cineel
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Robert Bassett
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    Did your professor intend for her remarks to be applied to finite fields? – Gerry Myerson Mar 17 '13 at 23:03
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    On $\Bbb F_2^2$ the product is not an inner product, as it is not positive-definite. (Positive definite implies for instance that $\langle x, x\rangle = 0 \Leftrightarrow x = 0$) – Arthur Mar 17 '13 at 23:05
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    Inner products are not really defined for finite fields. You need some form of order for positivity to work and that immediately requires characteristic $0$. – EuYu Mar 17 '13 at 23:05
  • You might find this useful: http://math.stackexchange.com/questions/185403/do-there-exist-vector-spaces-over-a-finite-field-that-have-a-dot-product also this: http://math.stackexchange.com/questions/49348/inner-product-spaces-over-finite-fields – Josephine Moeller Mar 17 '13 at 23:07
  • Inner products need a real or complex base field. Complex conjugacy is right in the definition of an inner product. – Ragib Zaman Mar 17 '13 at 23:08
  • @Arthur That makes sense. Thank you. However, I have often heard of mathematicians talk about orthogonality over $F_2$ using this inner product. Exactly what conditions of an inner product are necessary to properly introduce orthogonality? – Robert Bassett Mar 17 '13 at 23:16
  • The conditions other than the one about being positive-definite. – Gerry Myerson Mar 18 '13 at 02:37

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An inner product on a real or complex vector space $V$ is a bilinear map $V\times V\to K$ (where $K$ is the ground field, either $\bf R$ or $\bf C$), that satisfies conjugate symmetry and positive-definiteness; for the second property to make sense we have to realize that $\langle x,x\rangle$ is real for all $x\in V$ (this follows from the first property actually), so it makes sense to say it is nonnegative.

The map $\|x\|=\langle x,x\rangle^{1/2}$ will in fact be a vector space norm, and this norm induces a metric via the formula $d(u,v)=\|u-v\|$. The nontrivial part of checking these facts is using Cauchy-Schwarz for establishing the triangle inequality (fix an orthogonal basis to do it in).

It is not possible to define an inner product on a vector space over a field of positive characteristic, by definition. It is, however, possible to define bilinear forms $(\cdot,\cdot):{\bf F}_q^n\times{\bf F}_q^n\to{\bf F}_q$, and two vectors are orthogonal with respect to it if $(a,b)=0$. In most cases of positive characteristic and dimension greater than one it is possible to find an $x$ which is orthogonal to itself under the usual coordinate-determined dot product (this is actually an interesting number-theoretic question: over which finite fields and numbers $n$ do there exist $n$ scalars not all zero whose squares sum to zero?)

It is also not possible to define a metric $X\times X\to {\bf F}$ where $\bf F$ is a field of positive characteristic: by definition in the first place it would have to take real values, but furthermore it could not satisfy the triangle inequality since there can be no ordering in positive characteristic.

anon
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