Stacks project -- Comments

#4223 on tag 07I0 by Dario Weißmann

Dario Weißmann — Sun, 19 May 2019 05:31:27 GMT

Concerning the definition of the maps $\nabla:M\otimes \Omega^i \to M\otimes \Omega^{i+1}$ :

Several $\otimes$ should be $\wedge$ , i.e., $\begin{align} \nabla(bm\otimes \omega) - \nabla(m\otimes b\omega) & = \nabla(bm)\wedge \omega + bm\otimes d\omega - \nabla(m)\wedge b\omega - m\otimes d(b\omega)\\ & = b\nabla(m)\wedge \omega + m\otimes db \wedge \omega + bm\otimes d\omega - b\nabla(m)\wedge \omega - bm\otimes d\omega - m\otimes db\wedge \omega = 0 \end{align}$

#4222 on tag 07HT by Dario Weißmann

Dario Weißmann — Sun, 19 May 2019 04:44:00 GMT

Concerning the second paragraph of the proof. I suggest replacing "We will show this by induction ...." (till the end of the paragraph) by the shorter (and easier) direct calculation: The claim is clear for $n=1$ . Assume $n>1$ . Note that $\delta_i(z-y)$ lies in $(K\cap J(1))^{[2]}$ for $i>1$ . Calculating modulo $K^2 + (K\cap J(1))^{[2]}$ we have $\delta_n(z)=\delta_n(z-y+y)=\sum_{i=0}^n \delta_i(z-y)\delta_{n-i}(y)=\delta_{n-1}(y)\delta_1(z-y)+\delta_n(y).$ The claim follows.

#4221 on tag 09YS by Pawan Mishra

Pawan Mishra — Sun, 19 May 2019 02:25:40 GMT

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#4219 on tag 07J3 by Dario Weißmann

Dario Weißmann — Fri, 17 May 2019 08:48:30 GMT

Typo in the definition of the multiplication: $f \omega_2'+f'\omega_2'$ has too many $'$ s

There is also an issue how the multiplication formula is displayed. It overlaps with the sidebar. Works fine in the pdf version though.

#4217 on tag 09ZC by Fabrice Orgogozo

Fabrice Orgogozo — Wed, 15 May 2019 09:45:28 GMT

(And I gave a short proof of the fact that I don't know how to count...)

#4216 on tag 09ZC by Fabrice Orgogozo

Fabrice Orgogozo — Wed, 15 May 2019 09:43:43 GMT

4th sentence of the proof: sch -> such.

#4215 on tag 09LL by Frank

Frank — Tue, 14 May 2019 03:25:53 GMT

It seems better to clarify the second property of graded tensor product: the $p,q$ in $x\in M^p,y\in N^q$ are different from those in the direct sum $\bigoplus_{p+q=n}$ . In other words, here $p+q$ is only assumed to be no greater than $n$ . It seems better to choose different names for them.

#4214 on tag 03BX by David Holmes

David Holmes — Tue, 14 May 2019 04:55:13 GMT

I came here to make the same comment as Wessel, but then I saw Wessel's comment, so I won't (maybe I just did...). So maybe count this as another vote for changing it??

#4213 on tag 09YB by 羽山籍真

羽山籍真 — Mon, 13 May 2019 02:27:20 GMT

For the last sentence of the proof, $V$ has been used to denote an open of $Y$ , so it's not a subspace of $Z$ indeed, maybe replace it by $V\prime$ ?

#4212 on tag 02L3 by Sean Cotner

Sean Cotner — Fri, 10 May 2019 01:43:39 GMT

Small point: after replacing S by f(X), you still use the assumption that S is affine. I think this can be fixed by instead base changing to an arbitrary affine open contained in f(X), after which the rest of the proof goes through.