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Fix $n$ and let $P^n$ be projective $n$-space. Let $S = k[x_0, \dots, x_n]$. Set $A^0 = \bigoplus_{d \ge 0} H^0(P^n, \mathcal{O}(d))$ and $A^n = \bigoplus_{d < -n} H^n(P^n, \mathcal{O}(d))$.

I have been told that it is a "well-known" result that $A^0 \oplus A^n$ has an $A_\infty$-algebra structure that extends the usual algebra structure on $A^0$ and so that using the higher multiplication maps the algebra is generated by $H^0(P^n, \mathcal{O}(1))$ and $H^n(P^n, \mathcal{O}(-n-1))$.

Question: Is there a reference or can someone sketch a proof of the statement above?

There is probably a more general statement involving a scheme $X$ and an ample line bundle $L$ on $X$, though I'd already be interested in this special case.

Some thoughts: I have tried to realize this by putting a dga structure on the standard Cech complex (as in the proof of Theorem III.5.1 of Hartshorne's Algebraic Geometry book) and it looks like it works, but there is a problem with grading shift. To be precise, we have a complex $C^\bullet$ so that $C^d = \bigoplus_{I \subset \{0,\dots,n\}} S[x_I^{-1}]$ where $x_I = x_{i_0}\cdots x_{i_d}$ and $I$ ranges over all subsets of size $d+1$. Then there is an obvious multiplication structure on $C^\bullet$ but with a shift: $C^d \otimes C^e \to C^{d+e+1}$.

So it doesn't quite work (and also shifting the indexing does not resolve the problem because then I would be removing the algebra structure on $A^0$) but it seems that this shift by 1 is a common occurrence (like the Whitehead product on the homotopy groups of a space).

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  • $\begingroup$ If you set $X^n=C^{n-1}$ the degrees fit i.e. $X$ is a graded algebra, so you just have to check the Leibniz rule. $\endgroup$ Nov 21, 2014 at 17:12
  • $\begingroup$ "shifting the indexing does not resolve the problem because then I would be removing the algebra structure on $A^0$" $\endgroup$
    – Steven Sam
    Nov 21, 2014 at 19:24
  • $\begingroup$ So what is the product you define on $C$? If it doesn't give you the piece of A-infinity algebra structure you want to have... is it even associative? $\endgroup$ Nov 21, 2014 at 23:08
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    $\begingroup$ This paper may be of interest. arxiv.org/abs/math/0302178 $\endgroup$
    – user36931
    Aug 28, 2015 at 18:16

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