MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a number field with a fixed embedding $F \hookrightarrow \mathbb{C}$ such that the restriction of complex conjugation from $\mathbb{C}$ to $F$ is in Gal$(F/\mathbb{Q})$ and fix a Hermitian inner product $\langle v,w \rangle = \overline{v_1}w_1 + \cdots + \overline{v_n}w_n$ on $\mathbb{C}^n$ (with respect to the standard basis of $\mathbb{C}^n$. In particular, this restricts to a Hermitian inner product on $F^n$.

Now suppose we are given a unitary matrix $U$ on $F^n$ with respect to that inner product. It is well known (independent of unitarity) that $U$ is diagonalizable over some extension $E/F$ - for instance, take $E$ to be a splitting field of the minimal polynomial of $U$. This means there is a matrix $M$ over $E$ such that $M U M^{-1}$ is diagonal in the standard basis.

What if we instead want a unitary $W$ such that $W U W^\dagger$ is diagonal? This can be accomplished by working over a bigger extension $E'/E$ that includes some extra square roots of elements of E. Namely, given any $M$ that diagonalizes $U$ over $E$, just add in the square roots of the eigenvalues of $M^\dagger M$.

Now for my question:

Is there any sort of intrisic (i.e. independent of a choice of $M$) understanding of the extension $E'$? By understanding, I mean things like: is there a nice way of describing its generators over E? Can anything be said about its Galois group in general? When is it a semidirect product?

My actual interest is in the case where $F$ is cyclotomic and $U$ has finite order (and thus has roots of unity as eigenvalues, so $E$ is another cyclotomic field). Any advice on what is known in this specific, or otherwise the general case, would be be much appreciated.

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.