7
$\begingroup$

The regulator of a number field is essentially the covolume of the unit group embedded into the vector space $\{(x_1, \ldots, x_{r+s}): \sum_i x_i=0\}$ under the log embedding: $$x \mapsto (\log |\sigma_1(x)|, \ldots \log |\sigma_r(x)|, \log |\sigma_{r+1}(x)|^2, \ldots , \log |\sigma_{r+s}(x)|^2).$$

But you need to take some care in terms of what the right Euclidean structure is for defining volumes in $\{(x_1, \ldots, x_n): \sum_i x_i = 0\}$. What you're supposed to do is to use the Euclidean measure induced by any coordinate projection $\{(x_1, \ldots, x_{r+s}): \sum_i x_i = 0\} \rightarrow \mathbb{R}^{r+s-1}$. Equivalently, you could use the subspace Euclidean measure, but then normalize by dividing by $\sqrt{r+s}$.

My question is why is this the best normalization?

So far the only answer I have is that for the quadratic real case this makes the regulator exactly the log of the fundamental unit, which seems a very sensible convention. I guess it also makes the analytic class number formula slightly cleaner, but it's not obvious to me why this normalization exactly comes into the class number calculation.

My apologies if this question is too elementary.

$\endgroup$
4
  • 6
    $\begingroup$ I wouldn't think of it in terms of Euclidean measures. Given a volume on a vector space $V$, and a subspace $W$ of $V$, there is no natural notion of a volume restricted to $V$. However, if you have a natural volume on $V/W$, you can define a volume on $W$ by division. Here $V/W$ is isomorphic to $\mathbb R$ by the norm map (= sum of all the logarithms) and you divide by the standard measure on this $\mathbb R$. The importance of the norm map to the zeta function and thus to the class number formula should be clear... $\endgroup$
    – Will Sawin
    Nov 2 '20 at 15:59
  • $\begingroup$ Ah great, thanks! By division, you mean pick any additional vector in V and then compute the volume of the new parallelepiped in V divided by the length of the image of the new vector in V/W? $\endgroup$ Nov 2 '20 at 16:14
  • 1
    $\begingroup$ In terms of the determinant $\operatorname{det}(V) = \Lambda^{\textrm{max}} V$, we have $\operatorname{det}(V) \cong \operatorname{det}(W) \otimes \operatorname{det}(V/W)$ (there may be a choice of sign, which doesn't matter here). $\endgroup$ Nov 2 '20 at 19:13
  • $\begingroup$ If either of you want to write an answer I'll accept it, otherwise I'll write up the answer in comments myself when I get a minute. $\endgroup$ Nov 2 '20 at 21:29
4
$\begingroup$

Let $V$ be a vector space and $W$ a subspace.

Given a volume on $V$ and a volume on $V/W$, we can define a volume on $W$. (Here "volume" = "translation-invariant measure" but no real measure theory is being used. We could also define a volume as an element of the top wedge power of the dual vector space.)

Geometrically, this says that given a basis $v_1,\dotsc, v_m$ of $W$ which extends to a basis $v_1,\dotsc, v_n$ of $V$, the volume of the parallelepiped spanned by $v_1,\dotsc, v_m$ in $W$ is equal to the volume of the parallelepiped spanned by $v_1,\dotsc, v_n$ in $V$ times the volume of the parallelepiped spanned by $v_{m+1},\dotsc,v_n$ in $V/W$. (The same thing works with any shape in $V$ whose fibers under the projection map to $V/W$ are all translates of a fixed shape in $W$.)

We can express this algebraically with top forms as well, as François Brunault explains in the comments.

For $V$ the vector space of tuples $(x_1,\dotsc, x_{r+s})$, and $W$ the subspace $\sum_i x_i=0$, it is natural to identify $V/W$ with $\mathbb R$ under the map $\sum_i x_i$ and take the standard volume on $\mathbb R$. When $x_i$ are the logs of the individual coordinates, this identification corresponds to the norm map.

The reason this is the most commonly used normalization in number theory problem has to do with the fact that the norm map is used often in number theory. In particular, the norm is used in defining the zeta function, explaining why this approach gives nice formulas for the residue of the zeta function.

$\endgroup$
2
  • $\begingroup$ I think you switched partway through which space was $V$ and which one was $W$, leading to statements about $W/V$ and the volume of subsets of $W$ being quotients of the volumes of subsets of $V$. I edited in a way that I think preserved meaning while correcting these trivial errors. $\endgroup$
    – LSpice
    Nov 2 '20 at 22:46
  • 1
    $\begingroup$ @LSpice Thanks! $\endgroup$
    – Will Sawin
    Nov 2 '20 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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