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My question is whether the construction of higher Witt groups of a scheme in Schlichting's Hermitian K-theory of Exact Categories agrees with the definition in Balmer's chapter in the Handbook of K-theory when 2 is inverted (i.e. the schemes are over $\mathbb{Z}[\frac{1}{2}]$, although I'd also be happy to know whether this is true when we tensor the Witt groups with $\mathbb{Z}[\frac{1}{2}]$). See below for both definitions of higher Witt groups. When 2 is not inverted, they only necessarily agree on the 0-th Witt group.

More background if necessary

Schlichting defines higher Witt groups of an exact category with duality $(\mathcal{E}, *, \eta)$ as the homotopy groups of the geometric realization of a Hermitian Q construction, which is roughly as follows

Given an exact category $\mathcal{E}$ with duality, define the category $Q^h\mathcal{E}$ as the category where

  • objects are symmetric spaces in $\mathcal{E}$, i.e. pairs $(X,\varphi)$ of $X\in \text{ob }\mathcal{E}$ and an isomorphism $\varphi: X\mapsto X^{*}$ such that $\varphi^*\eta_X = \varphi$
  • morphisms from $(X, \varphi)$ to $(Y, \psi)$ are equivalence classes of diagrams $X \xleftarrow{p}U\xrightarrow{i}Y$ where $p$ is an admissible epimorphism, $i$ an admissible monomorphism, and the restrictions of the symmetric forms on $X$ and $Y$ to $U$ agree.
  • composition is pullback

(For a more precise statement, see Definition 4.1 in p.12 of Schlichting)

The $0$-th Witt group $\pi_0(|Q^h\mathcal{E}|)$ turns out to be the usual $$W_0(\mathcal{E}) = \{\text{isoclasses of symmetric spaces}\}/\{\text{metabolics}\}$$ (metabolics are defined in Schlichting, p. 6, Def 2.5, the 0th Witt group is defined in Schlichting p. 7, first paragraph of 2.2 and also in Balmer p. 7 Definition 1.1.27)

We can take Witt groups of a scheme $X$ by using the exact category of vector bundles over $X$ with the usual duality.

On the other hand, Balmer defines higher Witt groups of a triangulated category $\mathcal{K}$ with duality by taking $$W_0(\mathcal{K}) = \{\text{isoclasses of symmetric spaces}\}/\{\text{metabolics}\}$$ as above, and taking $$W_n(\mathcal{K}) = W_0(T^n\mathcal{K})$$ where $T^n$ denotes "applying the shift functor $n$ times" in the triangulated category, which can change the duality (see Balmer Def 1.4.1 and Def 1.4.4 for details). Now we can take Balmer Witt groups of a scheme $X$ by using the category of perfect complexes over $X$.

These definitions certainly agree on $W_0$, but they don't agree in the higher Witt groups (as can be seen in Remark 4.2 of Schlichting), and I want to know whether they do agree when $2$ is inverted as I said above. Thanks!

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No, the definition in Schlichting's first paper are not the "correct" definition of higher Witt groups (in any case they are not the analogue of Balmer's Witt groups), rather they are some shifted higher Grothendieck-Witt groups. He provided later a different definition that does coincide with Balmer's (but which is defined only when 2 is invertible).

For brevity let me refer to the following three papers by Marco Schlichting:

[Sch1] Hermitian K-theory of Exact Categories

[Sch2] The Mayer-Vietoris principle for Grothendieck-Witt groups of schemes

[Sch3] Hermitian K-theory, derived equivalences and Karoubi's fundamental theorem

In [Sch3] definition of the Grothendieck-Witt spectrum is provided when 2 is invertible. From this in [Prop. 7.2, Sch3] he deduces a model for Balmer's higher Witt groups as homotopy groups of the L-theory spectrum.

However, by using Proposition 6 in [Sch2] we see that the space $|Q^h\mathcal{E}|$ used in [Sch1] to define the Witt groups is not the 0-th space of the underlying spectrum. Rather it is the first space of the shifted Grothendieck-Witt spectrum $\operatorname{GW}^{[-1]}(\mathcal{E})$. Therefore we have $$\pi_i |Q^h\mathcal{E}|\cong \pi_{i-1}\operatorname{GW}^{[-1]}(\mathcal{E})$$

In particular its homotopy groups are not 4-periodic, and in general are not Witt groups.

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  • $\begingroup$ Maybe I should add that I do not know any definition of higher Witt groups that do not pass through some version of chain complexes, especially when 2 is not invertible (otherwise you can cheat via periodicity and use simple formulas in low degrees via formations, which are basically length one chain complexes) In fact if you know one I'd be rather interested in it... $\endgroup$ – Denis Nardin Jan 21 at 23:10
  • $\begingroup$ Thank you Denis, this is perfect! $\endgroup$ – Natalia Jan 21 at 23:10
  • $\begingroup$ Unfortunately, I'm new to this subject so I don't know either... :( $\endgroup$ – Natalia Jan 21 at 23:13

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