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Let $X$ be a smooth projective toric variety. The Chern character gives an isomorphism of rings: $$\operatorname{Ch}:K_{0}(X)\otimes\mathbb{Q} \to A(X)\otimes \mathbb{Q} $$ where $K_{0}(X)$ is the Grothendieck group of vector bundles on $X$ and $A(X)$ is the Chow ring of $X$. This map seems only well-defined over $\mathbb{Q}$, but I was wondering (likely naively) if there is possibly an integral isomorphism (i.e. without tensor with $\mathbb{Q}$)?

Why might we hope for such a map? Fulton and Strumfels showed that there exists an isomorphism $\mathcal{D}_{X}:A^{k}(X)\to \operatorname{Hom}(A_{K}(X),\mathbb{Z})$ where $A^{k}(X)$ and $A_{k}(X)$ are the Chow cohomology and homology groups respectively. In particular, this means that the Chow ring $A(X)$ of a smooth toric variety is torsion free. In the couple very (very) simple examples I've done $K_{0}(X)$ also seems torsion free, although I am unsure whether this is true generally.

Of course, even if both $K_{0}(X)$ and $A(X)$ are torsion free there need not be an isomorphism between them, but one can hope.

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    $\begingroup$ Welcome to Mathoverflow! Are you asking about isomorphism as abelian groups or as rings? I think these groups will be free of the same rank (equal to the number of maximal cones in the fan) but not in general isomorphic as rings. $\endgroup$
    – Evgeny Shinder
    Jun 8, 2021 at 22:36
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    $\begingroup$ @EvgenyShinder I was hoping for them to be isomorphic as rings, but I would also be interested to know that they are isomorphic as abelian groups. :) Why are they both free of the same rank? $\endgroup$
    – Juliette Bruce
    Jun 8, 2021 at 23:02
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    $\begingroup$ I think that they are free of the same rank is covered in (the proof of) Lemma 1 of Morelli's "The K-theory of a Toric Variety". $\endgroup$
    – Hunter Spink
    Jun 9, 2021 at 0:23
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    $\begingroup$ For surfaces, I think they are isomorphic as rings for a dumb reason. Choose any $\mathbb{Z}$-basis $L_1$, $L_2$, ..., $L_k$ for $\mathrm{Pic}(X)$. Then a basis for $K_0$ is $1$, $L_1-1$, $L_2-1$, ..., $L_k-1$ and $p$, where $p$ is the structure sheaf of a point. Letting $D_j \in A^1(X)$ be $c_1(L_j)$ and letting $q$ be the class of a point in $A^2$, send $L_j-1$ to $D_j$ and send $p$ to $q$. I believe this respects multiplication. It is completely noncanonical though: Take a different basis for $\mathrm{Pic}(X)$ and you get a different map. $\endgroup$
    – David E Speyer
    Jun 9, 2021 at 2:04
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    $\begingroup$ By the way, if anyone answers this question, it would be nice if they would also address the associated graded ring $\mathrm{gr}\ K^0$. $\endgroup$
    – David E Speyer
    Jun 9, 2021 at 12:55

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For the permutahedral toric variety associated to the type A hyperplane arrangement, there is such an isomorphism -- see Section 10 of Berget-Eur-Spink-Tseng https://arxiv.org/abs/2103.08021 . I believe this isomorphism also descends to toric varieties with coarser fans, and many small examples (like e.g. projective space) are special cases of this.

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