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Stream: theory: mathematics

Topic: tangent space of power object in smooth sets

Matteo Capucci (he/him) (May 07 2024 at 10:05):

Has anyone ever worked out how the tangent space of a power object in $\bf Sh(Cart)$ looks like?

Matteo Capucci (he/him) (May 07 2024 at 10:35):

Actually, even understanding what the power object of a cartesian space $\R^n$ is is a bit of an head-scratcher for me

Matteo Capucci (he/him) (May 07 2024 at 10:37):

If I'm not mistaken, it should be the sheaf

$$X \mapsto P\R^n(X) = \Omega^{\R^n}(X) = \int_Y {\bf Cart}(Y,X) \times \Omega(Y) \to {\bf Cart}(Y,\R^n)$$

Matteo Capucci (he/him) (May 07 2024 at 10:44):

and I'm not sure how to think about it

David Corfield (May 07 2024 at 14:21):

You might also pose this question here.

Kevin Carlson (May 07 2024 at 17:55):

I'm a bit confused about your end expression, but $\Omega^{\mathbb R^n}(X)=\mathbf{Sh}(\mathbf{Cart})(X\times \mathbb R^n,\Omega)=\Omega(X\times \mathbb R^n)$ should be the set of closed sieves on $X\times \mathbb R^n,$ no?

Matteo Capucci (he/him) (May 07 2024 at 20:16):

Kevin Carlson said:

I'm a bit confused about your end expression, but $\Omega^{\mathbb R^n}(X)=\mathbf{Sh}(\mathbf{Cart})(X\times \mathbb R^n,\Omega)=\Omega(X\times \mathbb R^n)$ should be the set of closed sieves on $X\times \mathbb R^n,$ no?

🤦🏻‍♂️ I switched base and exponent in the formula... Then of course it reduces to what you say.

Matteo Capucci (he/him) (May 07 2024 at 20:16):

Thanks