As the title states, I am looking for some examples of non trivial albeit elementary calculations of algebraic hermitian and non-hermitian $K$-theory groups.
Concretely, I would be interested in having some hands-on idempotent/"by definition" oriented calculations as opposed to homotopy theoretic ones, since I've not thoroughly studied higher $K$-theory nor spectra in detail yet.
Some examples for algebraic non-hermitian $K$-theory that fit this criteria would be:
- $K_0(R) = \mathbb{Z}$ when $R$ is a PID or a local ring.
- If $R$ is a Dedekind ring, the reduced $K_0$ is isomorphic to its class group.
- $K_0$ is cocontinuous and Morita invariant. This gives a somewhat explicit description of $K_0$ for ultramatricial algebras.
As for algebraic hermitian $K$-theory, I only know of the definition itself, but If I am not mistaken, we should have $K^h_0(k) = \mathbb{Z}$ for any field equipped with any involution (since any finitely dimensional inner-product vector space over $k$ has an orthonormal basis, it must be isomorphic to $k^n$ with the trivial inner product induced by the involution of $k$, right?).
For example, is there a straighforward way to compute $K^h_0(\mathbb{Z})$? Here I mean $\mathbb{Z}$ equipped with its trivial involution. I would expect this to differ from non-hermitian $K_0$, as inner-product modules over $\mathbb{Z}$ look (to me, at least) much harder to classify.
Edit: okay, so, my guess for $K_0^h(k)$ was way off the mark. I'd appreciate a sanity check on the following calculations.
Existence of an orthonormal basis (which iirc may also depend on the characteristic of the field) for any vector spaces only tells us that a module equipped with a hermitian form is equivalent to $(k^n,A)$ for some hermitian $A \in \mathsf{M}_n(k)$. Two such modules $(k^n,A)$ and $(k^m,B)$ will be isomorphic iff $n = m$ and $A = SBS^\ast$ for some invertible matrix $S$.
When $k = \mathbb{R}$, we know that any symmetric form is determined by its signature. This gives a map $$(k,l) \in \mathbb{N}_0^2 \mapsto (\mathbb{k}^{k+l}, \mathsf{diag}(\overbrace{1,\ldots, 1}^k,\overbrace{-1,\ldots, -1}^l)$$ which is an isomorphism of monoids, so $K_0^h(\mathbb{R}) = \mathbb{Z}^2. $ and
$$W(\mathbb{R}) = \mathrm{coker}( K_0(\mathbb{R}) \to K_0^h(\mathbb{R})) \simeq \mathrm{coker}(\mathbb{Z} \xrightarrow{(1,-1)}\mathbb{Z}^2) = \mathbb{Z}.$$
Likewise, when $k = \mathbb{C}$ any Hermitian matrix $A$ is unitarily diagonalizable with real eigenvalues. If moreover $A$ is invertible, its eigenvalues $d_1 \leq \ldots \leq d_n$ are non-zero. Thus, after a change of basis, we may assume $A = \mathsf{diag}(d_1, \ldots, d_n)$ is diagonal with non-zero diagonal entries. For some choice of complex square roots, taking $D = \mathsf{diag}(1/d_1^{1/2}, \ldots, 1/d_n^{1/2})$ gives
$$ D^tAD = I. $$
Hence $K_0^h(\mathbb{C}) = \mathbb{Z}$, generated by $(\mathbb{C},1)$, and since the class of hyperbolic space $(\mathbb{C}^2, \mathsf{diag}(1,-1))$ is $[(\mathbb{C}^2,I_2)] = [(\mathbb{C},1)]^2$ the Witt group of $\mathbb{C}$ is
$$ W(\mathbb{C}) = K_0^h(\mathbb{C})/\langle [(\mathbb{C},1)]^2\rangle = \mathbb{Z}_2. $$