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Derived geometry explains how to remove the transversality condition and make sense out of a nontransversal intersection. For example, if $X$ and $Y$ are embedded submanifolds of a manifold (or spaceform) $Z$, then the intersection of $X$ and $Y$ is a derived manifold of dimension $\dim X+\dim Y-\dim Z$, which can be negative.

For a derived manifold of dimension $-2$, how I can write its metric? Can someone give me an example of such a metric?

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    $\begingroup$ You could try to define Riemannian metrics for derived manifolds in terms of the cotangent complex $L_X$, but non-degeneracy will only be possible when the cotangent complex is concentrated in degree $0$, thus excluding manifolds of negative dimension. However, under some circumstances, some form of $n$-shifted Riemannian metric may be possible, i.e. $L_X^* \simeq L_X[n]$. $\endgroup$
    – Jon Pridham
    Commented Mar 2, 2019 at 9:17
  • $\begingroup$ Thank you very much for your comment. I don't know how to get a metric for manifold with negative dimension $\endgroup$
    – MathDG
    Commented Mar 2, 2019 at 11:14

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As far as I am aware, there is nothing in the literature that treats Riemannian or pseudo-Riemannian metrics on derived smooth manifolds. However, there is an extensive treatment of symplectic structures on derived stacks by Pantev, Toën, Vaquié, and Vezzosi: Shifted symplectic structures.

Due to the Koszul sign rule, a pseudosymplectic structure (i.e., a symplectic structure without the integrability condition) is nothing else than a pseudo-Riemannian metric on the cotangent complex shifted in degree by 1.

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  • $\begingroup$ Dear prof. Pavlov, thank you very much for your help. So I can consider a -2-dimensional derived-manifold as a derived manifold endowed with -2-shifted symplectic structure, is correct? Then like a metric for a -2-dimensional derived-manifold, can I consider a 2-symplectic structure metric on a derived stack? $\endgroup$
    – MathDG
    Commented Mar 2, 2019 at 8:29
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    $\begingroup$ A symplectic form has more restrictions than a shifted pseudo-Riemannian metric, because of the closure conditions or closing structures. For shifted symplectic structures in the smooth setting, see arXiv:1804.07622.In answer to the OP's comment, n-shifted and n-dimensional are very different. $\endgroup$
    – Jon Pridham
    Commented Mar 2, 2019 at 9:10
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    $\begingroup$ @OP u would do well to be a bit clearer re what u want here. As Pridham mentions there's a shift involved, a good basic example would be the odd line over the reals. The simplest care should be real derived vec spaces (chain complexes that is) with non deg shifted symm forms, nb that unlike in the case of non derived vec spaces such a form needn't exist (it implies a certain symmetry on the cohom groups) $\endgroup$
    – user108998
    Commented Mar 2, 2019 at 20:01
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    $\begingroup$ @exxxit8: If points are understood in the usual sense, than most of the structure of a derived manifold is irrelevant: points in M are morphisms pt→M, which translates to a morphism of C^∞-dgas C^∞(M)→C^∞(pt)=**R**, which is the same thing as π_0(C^∞(M))→**R**, i.e., a point in the underlying ordinary smooth space of M. $\endgroup$
    – Dmitri Pavlov
    Commented Mar 3, 2019 at 19:07
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    $\begingroup$ @exxxit8: Yes. The derived part can be thought of as an infinitesimal thickening of some kind, so it does not affect distances between ordinary points. $\endgroup$
    – Dmitri Pavlov
    Commented Mar 4, 2019 at 17:09

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