Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How is the trace pairing function $(x,y) \mapsto Tr(xy)$ on a number field an analogue of the dot product in euclidean space?

(This is a view shared by Keith Conrad and can be found in his notes Discriminants... and The Different Ideal)

share|improve this question
It's bilinear... it's nondegenerate... –  Qiaochu Yuan May 26 '12 at 3:51
Section 3 of the 2nd handout you link to sets up the similarity between lattices in Euclidean space and lattices in number fields. I'm not sure why you are asking after reading that "how" there is an analogue. Did it not come out in Section 3? Beyond the setting of number fields, for any finite separable extension of fields $L/K$, the trace pairing $L \times L \rightarrow K$ is perfect, so every element of the $K$-dual space of $L$ has the form $f(x) = {\rm Tr}_{L/K}(xy)$ for a unique $y \in L$. Likewise, element of the dual space of ${\mathbf R}^n$ is $f(v) = v\cdot w$ for a unique $w$. –  KCd Jun 2 '12 at 7:10

1 Answer 1

The vector space of $n$ by $n$ matrices with real entries has a positive definite inner product given in the convenient shape $$ \langle A, B \rangle = \; \mbox{tr} \; \left(A B^T \right) $$ which is pretty much what I am seeing in his notes.

Multiplication in a field is linear over the base field, so you get an analogy. Depending how you want to order things, let the $e_i$ be a basis for your field over the base field. Then multiplication by some field element $x$ is completely determined by the matrix of values $x_{ij}$ such that $$ x e_i = \sum x_{ij} e_j. $$ So there is your matrix.

I've got to say, for utter improvisation, this is pretty good.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.