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Let MHS be the category of rational mixed Hodge structures. In particular, it contains extensions of Tate objects $\mathbb{Q}(n)$ for each integer $n$. Here $\mathbb{Q}(n)$ is the only one dimensional $\mathbb{Q}$-pure Hodge structure of weight $-2n$.

Can someone indicate me how to compute the group of extensions

$Ext^1_{MHS}(\mathbb{Q}(n), \mathbb{Q}(m))$?

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To unpack and correct Dan's answer:

The extensions $\mathrm{Ext}^1(A,B)$ for $A$ and $B$ mixed Hodge structures are given by considering the direct sum $C_{\mathbb{Q}}=A_{\mathbb{Q}}\oplus B_{\mathbb{Q}}$ with the obvious weight filtration $W_mC_{\mathbb{Q}}=W_mA_{\mathbb{Q}}\oplus W_mB_{\mathbb{Q}}$ , and choosing the Hodge filtration $F_kC=\{(a,b+\varphi(a))\in C_{\mathbb{C}}|a\in A_{\mathbb{C}},b\in B_{\mathbb{C}}\}$ for $\varphi \colon A\to B$ a map preserving the weight filtration. The resulting extension is trivial if and only if $\varphi=\varphi'+\varphi''$ is the sum of a map $\varphi'$ preserving the Hodge filtration, and a map $\varphi''\colon A_{\mathbb{Q}}\to B_{\mathbb{Q}}$ which is defined rationally (if you want integral or real MHS, just replace $\mathbb{Q}$ by $\mathbb{Z}$ or $\mathbb{R}$ in the definition of $\varphi''$.

In the specific case of $\mathrm{Ext}^1(\mathbb{Q}(n),\mathbb{Q}(m))$, we have:

  • if $m<n$, then only the trivial map preserves the weight filtration, so the Ext group is 0.
  • if $m=n$, then any map preserves the weight filtration, but it also preserves the Hodge filtration, and the extensions are trivial.
  • if $m> n$, then any map preserves the weight filtration, but only the trivial preserves the Hodge filtration. So we end up with $\mathrm{Ext}^1(\mathbb{Q}(n),\mathbb{Q}(m))=\mathrm{Hom}_{\mathbb C}(\mathbb C,\mathbb C)/\mathrm{Hom}_{\mathbb Q}(\mathbb Q,\mathbb Q)$, so it's isomorphic to $\mathbb{C}/\mathbb{Q}$, though I think people like to normalize to $\mathbb{C}/(2\pi i)^{m-n}\mathbb{Q}$ (this just depends on whether you choose the generator of $\mathrm{Hom}_{\mathbb C}(\mathbb C,\mathbb C)$ to preserve the rational structures and thus send $(2\pi i)^n$ to $(2\pi i)^m$, or be the identity map).

Dan's answer was assuming you want integral Hodge structures, so you get $\mathbb{C}/\mathbb{Z}\cong \mathbb{C}^*$ instead.

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This extension group is $\mathbf C^\ast$ if $n < m$ and $0$ otherwise. See e.g. Carlson, "Extensions of mixed Hodge structures".

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  • $\begingroup$ I'm trying to read the paper but I find it hard. Can you give some help/intuition of why this result is true? $\endgroup$
    – MHS93
    May 26 '14 at 10:54
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    $\begingroup$ This answer is assuming integral mixed Hodge structures. Rational points on the unit circle in $\mathbb{C}^*$ give MHS which are rationally trivial, but not integrally. $\endgroup$
    – Ben Webster
    Oct 13 at 12:50
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    $\begingroup$ Thanks for the correction, Ben. $\endgroup$ Oct 13 at 13:20
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    $\begingroup$ No problem (and realized rereading I was a little abrupt in my first comment; sorry). I just spent a while being confused by the same paper, so figured I would do something constructive with it. $\endgroup$
    – Ben Webster
    Oct 13 at 14:17

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