In Deligne's 1972 article on the Weil conjecture for K3 surfaces, he essentially constructed an inclusion of Shimura data $(GSpin(V),X)\subset(GSp(W),H(W))$, where $V$ is a vector space over $\mathbb{Q}$ of dimension $n+2$ endowed with a quadratic form $q$ of signature $(n,2)$, $C(V)$ the Clifford algebra associated to $(V,q)$, $C^+(V)$ the even part of $C(V)$, $GSpin(V)$ the reductive $\mathbb{Q}$-group of invertible elements in $C^+(V)$ that preserve $V$ under conjugation in $C(V)$. The Hermitian symmetric domain $X$ can be identified as the space of $q$-isotropic negative planes in $V_\mathbb{R}$, as is described in Kudla's article "algebraic cycles in Shimura varieties of orthogonal type". The inclusion of Shimura data mentioned above is given by the representation $(W,\rho_W)$ by left translation of $GSpin(V)\subset C^+(V)$ on $W=C^+(V)$, which respects a canonically defined symplectic pairing up to scalar. Finally $H(W)$ is the Siegel double space associated to the above symplectic structure.

Note that for any $x\in X$ (or in $H(W)$), $(W,\rho\circ x)$ is a Hodge structure of type $(-1,0),(0,-1)$, hence $W\otimes W$ underlies a Hodge structure of type $(-2,0),(-1,-1),(0,-2)$. On the other hand, the canonical representation $(V,\rho_V)$ of $GSpin(V)$ gives Hodge structures $(V,\rho_V\circ x)$ of type $(-2,0),(-1,-1),(0,-2)$.

My question : is there a natural embedding of $V$ into $W\otimes W$ as a subrepresentation of $GSpin(V)$? if there is, then I can naturally understand $V$ as a Hodge substructure of $W\otimes W$. The Hodge types already coincile, but I don't know if it follows from some simple universal construction.

  • $\begingroup$ note that in Deligne's 1972 article he hasn't modified yet the sigh rules as his later articles on Shimura varieties and Hodge structures. He showed that for the canonical representation of $SO(V)$ on $V$, one can find Hodge structure of type $(-1,1),(0,0),(1,-1)$ on $V$ through the Shimura datum $(SO(V),X')$, the later being deduced from $(GSpin(V),X)$ by taking adjoint group. If one considers the action of $GSpin(V)$ on $V$, one finds a non-trivial action of the center $\mathbb{G}_m$ on $V$, and simple calculations shows that in this case the Hodge types are shifted by $(-1,-1)$. $\endgroup$ – genshin Dec 17 '10 at 11:31
  • $\begingroup$ sign rules instead of sigh rules... $\endgroup$ – genshin Dec 17 '10 at 11:34
  • $\begingroup$ I have a related question. Is it possible to characterise the image of V inside W\otimes W ? $\endgroup$ – Rogelio Yoyontzin Nov 10 '11 at 5:11


Indeed, there does exist an embedding of Hodge structures as you mention. The thing works as follows.

Fix an element $v_0\in V$. Then $V$ acts on the vector space $C^+(V)$ by $v\mapsto(x\mapsto vxv_0)$. This induces an embedding $V \hookrightarrow End_k(C^+(V)$. This is equivariant with respect to the actions of $CSpin(V)$, where $CSpin(V)$ acts on $V$ by conjugation, hence by definition of the Hodge structures, it gives a morphism of weight zero Hodge structures $V(1) \hookrightarrow End_k(C^+(V)$.

Now a polarization of $C^+(V)$ identifies its dual with $C^+(V)(-1)$, hence a embedding of weight $2$ Hodge structures $$V\hookrightarrow C^+(V)\otimes C^+(V).$$

All the best

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    $\begingroup$ Thanks a great deal, but i'm a little confused. It is clear that $V$ acts on the total spinor algebra $C(V)$ by multiplication on the left. But how should this preserve the grading? $V$ is contained in the odd part $C^_(V)$ while $C^+(V)$ is even. Do I need to find a natural identification of $End(C^+(V))$ with something else? Maybe I misunderstood your explanations. $\endgroup$ – genshin Dec 18 '10 at 0:15
  • $\begingroup$ Sorry for the mistake, and thanks for pointing it out ! Hopefully the answer is closer to correct now. $\endgroup$ – F C Dec 18 '10 at 8:43
  • $\begingroup$ Dear F C, I am probably just confused, but don't we need some property of $v_0$ for this to work as written? (Because if $v_0$ is random, then after conjugating on the RHS by an element $u$ in $CSpin(V)$, we would obtain not the morphism $x \mapsto uvu^{-1} x v_0,$ but rather $x \mapsto u v u^{-1} x u v_0 u^{-1}$.) $\endgroup$ – Emerton Dec 18 '10 at 9:07
  • $\begingroup$ Dear Emerton, the thing is that the action of $CSpin(V)$ on $C^+(V)$ that you want to consider to get the weight one HS is by multiplication on the left, and not by conjugation. The action on $End_k(C^+(V))$ is thus given by $(u.f)(x)=uf(u^{-1}x)$, which gives the correct formula. $\endgroup$ – F C Dec 18 '10 at 9:15

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