Let $F$ be a number field with ring of integers $\mathfrak{o}$. Let $(V,Q)$ be a quadratic space of dimension $n$ over $F$, and let $L$ be a free lattice in $V$ (i.e. $L\cong\mathfrak{o}^n$). If the class number of $F$ is odd, then the genus of $L$ consists of free lattices, because the squares of their Steinitz classes are equal to the class of the volume of $L$, i.e. $1$. Also, in this case, the genus of $L$ consists of a single spinor genus.

**Question 1.** When is it true that the genus of $L$ consists of free lattices?

**Question 2.** When is it true that the spinor genus of $L$ consists of free lattices?

The questions above and the ones below are motivated by comparing the mass formula in its original form proved by Siegel, and in its adelic form developed by Tamagawa, Weil, etc. Assume $F$ is totally real and $Q$ is totally positive definite. The formula in its original form concerns the average number of representations of an integer $a\in\mathfrak{o}$ by the classes of quadratic forms over $\mathfrak{o}$ in the genus of $Q$. The theorems of Hasse-Minkowski and Witt allow one to reformulate this average as one over a genus of classes of representations $(x,M)$, where $x\in V$ is fixed with $Q(x)=a$, and $M$ is any *free lattice* in the genus of $L$ that contains $x$. In the adelic form it is no longer required that $M$ be free.

**Question 3.** Is there a simple reason why the classical and adelic formulations give the same answer? It seems that Siegel averages over fewer representations $(x,M)$ than Tamagawa, Weil.

**Question 4.** Is there a classical notion of the spinor genus of $Q$?

I would appreciate any thoughts, results, or pointers to the literature.

**Added 1.** It seems that the answer to Questions 1-2 are "always true" and this also gives the answer to Questions 3-4: "there is really no difference between the classical and adelic notions in the present setting". Namely, by this paper (based on a method of Kneser from 1957) each class in the genus of $L$ can be reached from the class of $L$ by successively taking $\mathfrak{p}$-neighbors for two suitable prime ideals $\mathfrak{p}$ (which can be chosen to lie outside any prescribed finite set of prime ideals), and $\mathfrak{p}$-neighbors always have the same Steinitz class. I suspect that this conclusion can be reached more directly, but I am no expert, so I am still awaiting thoughts, results, or pointers to the literature.

**Added 2.** Inspired by Rainer Schulze-Pillot's answer below, here is a slightly simpler way to show that the Steinitz class of a lattice $L$ is determined by the volume of $L$, hence also by the genus of $L$. We can write $L$ as
$$ L=\mathfrak{a}_1x_1+\dots+\mathfrak{a}_nx_n, $$
where $(x_i)$ is a suitable basis of $V$, and the $\mathfrak{a}_i$'s are fractional ideals in $F$. By definition, the volume of $L$ is the fractional ideal
$$ \mathfrak{v}L = \mathfrak{a}_1^2\dots \mathfrak{a}_n^2\cdot d(x_1,\dots,x_r), $$
where $d(x_1,\dots,x_r)$ is the discriminant of the basis $(x_i)$. The latter equals the discriminant of $V$ modulo $(F^\times)^2$, hence $\mathfrak{vL}$ really determines $\mathfrak{a}_1\dots \mathfrak{a}_n$ modulo $F^\times$ (because the multiplicative group of fractional ideals is torsion free), which equals the Steinitz class of $L$.