# extensions of mixed Hodge structures

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

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

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 $$m, then only the trivial map preserves the weight filtration, so the Ext group is 0.
• if $$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.

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

• I'm trying to read the paper but I find it hard. Can you give some help/intuition of why this result is true? May 26 '14 at 10:54
• 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. Oct 13 at 12:50
• Thanks for the correction, Ben. Oct 13 at 13:20
• 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. Oct 13 at 14:17