One of the standard conjectures in algebraic geometry is that an operator $\Lambda$ on the cohomology algebra of a projective variety is algebraic. To my lying eyes it looks like there are two definitions of the operator $\Lambda$ out there and that the two don't agree.

On the algebraic side, let $X$ be a projective variety of dimension $n$, equipped with the first Chern class $w$ of an ample line bundle $A$. We then get a Lefschetz operator on cohomology by setting $L u = u \wedge w$ for a class $u \in H^*(X)$. Because of the Lefschetz theorems we can define another operator $\Lambda_a u = (L^{n-k+2})^{-1} \circ L \circ L^{n-k}$ on $k$-classes $u$ when $k \leq n$ (when $k > n$ there's a similar definition that doesn't matter here). By definition this operator satisfies $\Lambda_a L = \operatorname{id}$.

On the geometric side, let $X$ be a compact Kahler manifold of dimension $n$, equipped with a Kahler metric $\omega$. (If the cohomology class of $\omega$ is entire then it is the curvature form of an ample line bundle on $X$ and everything is algebraic.) As before we get a Lefschetz operator on forms and cohomology by setting $L_g u = u \wedge \omega$. The metric $\omega$ also defines a Hodge star operator $*$ that operates on forms, which defines an inner product on forms, and the adjoint of $L$ with respect to that inner product is $\Lambda_g$. The Kahler identities say (amongst other things) that $[L,\Lambda_g]\,u = (k-n) u$ for a $k$-form $u$. These operators descend to the cohomological level, either by representing cohomology classes by harmonic forms or by working out purely cohomological definitions of the operators from primitive decompositions of cohomology classes (that the two agree necessarily passes through harmonic representatives).

Now, to me these definitions do not agree. For one, the commutation identity $[L,\Lambda_g]\,u = (k-n) u$ on the geometric side is incompatible with $\Lambda_a L = \operatorname{id}$ on the algebraic side. I don't think this difference is just a matter of overcoming not having all of Hodge theory in arbitrary characteristic; if we have primitive decompositions of classes we can define the $*$ and $\Lambda$ operators just fine on cohomology without reference to differential forms, which I would think is what we'd want to do if we wanted to emulate Hodge theory on Kahler manifolds in the algebraic world. Apparently the people who defined these things (Grothendieck and co. I imagine) didn't agree.

Why not? Why is one of the standard conjectures stated in terms of an operator that doesn't seem to be used in the classical world?


In addition to Kleiman's "Algebraic cycles and the Weil conjectures" mentioned by abx, I can highly recommend reading the first article in the Motives proceedings (Jannsen/Kleiman/Serre) by Kleiman: "The standard conjectures".

It introduces several $\Lambda$ and $\star$ operators, and shows that when one is algebraic, then so are the others. Further, it shows implications to weaker conjectures, such as algebraicity of the Künneth projectors.


I wonder where you have found this definition of $\Lambda_a $. My standard (!) reference for the standard conjectures is Kleiman's "Algebraic cycles and the Weil conjectures", in "10 exposés sur la cohomologie des schémas". Here they define 2 closely related operators, named $\Lambda $ and ${}^c\Lambda$; both are defined using the Lefschetz decomposition (which you don't use to define your $\Lambda_a$). The latter one satisfies $[L,{}^c\Lambda]=(k-n)$ on $H^k$, so it is the exact analogue of your $\Lambda_g$.


I will show that $\Lambda_a$ is algebraic if and only if $\Lambda_g$ is. All cohomology groups are with rational coefficients.

Let $H^k_{prim}$ be the primitive cohomology in $H^k$. Then $$H^{\ast}(X) = \bigoplus_{j=0}^n \bigoplus_{r=0}^j L^r H^{n-j}_{prim}(X).$$ Multiplication by $L$ is an isomorphism $L^r H^{n-j}_{prim}(X) \to L^{r+1} H^{n-j}_{prim}(X)$ for $0 \leq r < j$, and we will choose bases so that this isomorphism is the identity.

Multiplication by the $\Lambda$'s takes $L^{r+1} H^{n-j}_{prim}(X)$ to $L^r H^{n-j}_{prim}(X)$. For $\Lambda_a$, this map is $\mathrm{Id}$; for $\Lambda_g$ this map is $(r+1) (j-r) \mathrm{Id}$.

Write $\pi(j,r)$ for the projection onto $L^r H^{n-j}_{prim}$. So $$\Lambda_g = \sum_{j,r} (r+1) (j-r) \ \Lambda_a \ \pi(j,r) \quad \Lambda_a = \sum_{j,r} \frac{1}{(r+1) (j-r)} \ \Lambda_g \ \pi(j,r).$$ We will show that $\pi(j,r)$ is in the $\mathbb{Q}$-algebra generated by $(L, \Lambda_{\bullet})$, for $\bullet$ either $a$ or $g$. Thus, the above expressions show that $\Lambda_g$ is in the $\mathbb{Q}$ generated by $\Lambda_a$ and vice-versa.

Let $V_j = \bigoplus_{r=0}^j L^r H^{n-j}_{prim}(X)$ and let $\pi(j)$ be the projection onto $V_j$.

Assuming $\Lambda_a$ is algebraic: Set $$K = \Lambda_a^n L^n + \Lambda_a^{n-1} L^{n-1} + \cdots + \Lambda_a L + 1 + L \Lambda_a + \cdots + L^n \Lambda_a^n.$$ I get that $K$ acts by $j+1$ on $V_j$. By Lagrange interpolation, there is a polynomial $f_j$ with $f_j(j+1)=1$ and $f_j(k+1)=0$ for $k$ between $0$ and $n$ other than $j$. So $\pi(j) = f_j(K)$. I get that $$\pi(j,r) = L^r \Lambda_a^j L^{j-r} f_j(K).$$ So $\pi(j,r)$ is in the ring generated by $L$ and $\Lambda_a$.

Assuming $\Lambda_g$ is algebraic: Set $\Omega = (L \Lambda_g - \Lambda_g L)^2 + 2 (L \Lambda_g + \Lambda_g L)+1$. If I didn't make any errors, $\Omega$ is the Casimir, and acts on $V_j$ by $(j+1)^2$. By Lagrange interpolation, there is a polynomial $g_j$ with rational coefficients so that $g_j((j+1)^2) = 1$ and $g_j((k+1)^2)=0$ for any $k$ between $0$ and $n$ other then $j$. So $\pi(j) = g_j(\Omega)$. Then I get that $$\pi(j,r)= \frac{1}{(j!)^2} L^r \Lambda_g^j L^{j-r} g_j(\Omega) .$$ So $\pi(j,r)$ is in the ring generated by $L$ and $\Lambda_g$.

  • $\begingroup$ Hmm... I can't get the commutation identity to work out with $\Lambda_g$ a constant multiple of $\Lambda_a$, but that could just be me. Do you have an idea of a reference for this equality? $\endgroup$ – Gunnar Þór Magnússon Jan 5 '14 at 20:14
  • $\begingroup$ Yup, the identity is messier than I had claimed; we need to know the length of the $\mathfrak{sl}_2$ string as well as the cohomology degree. See if the rewritten version works. $\endgroup$ – David E Speyer Jan 8 '14 at 16:17
  • $\begingroup$ Ahh, I didn't notice that jmc had already explained this. $\endgroup$ – David E Speyer Jan 8 '14 at 16:17

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