Let $f\colon X\subset \mathbb{P}^n$ be a complex projective subvariety. We define the ($p$th) Chow monoid of $X$ as $$\mathcal{C}_p(X,f)=\bigsqcup_{d\geq 0}\mathcal{C}_{p,d}(X,f),$$

where $\mathcal{C}_{p,d}(X,f)$ is the Chow variety parametrizing the $p$-dimensional cycles of degree $d\geq 0$ supported in $X$ and, by assumption, $\mathcal{C}_{p,0}(X,f)$ consists only of the cycle with empty support which is the neutal element in the addtion law of the Chow monoid. By puttion on each $\mathcal{C}_{p,d}(X,f)$ the analytic topology we endow $\mathcal{C}_p(X,f)$ with the disjoint union topology. I'm trying to check that the topology on $\mathcal{C}_{p}(X,f)$ does not depend from the choice of the embedding $f\colon X\subset\mathbb{P}^n$, but only from the isomorphism class of $X$. To make precise this statement usually one defines the so called algebraic maps.

Def.(Continuos Algebraic Map) A map $F\colon X\longrightarrow Y$ between irreducible projective variety is called continuos algebraic map (CAM) if its graph $\Gamma_F\subset X\times Y$ is closed.

The definition above is that given by H.B. Lawson in Spaces of Algebraic Cycles.

In the paper Introduction to Lawson Homology by C. Peters and S. Kosarew, the authors say that $F\colon X\longrightarrow Y$ is a CAM if in addition to the assumptions of the definition above the projection $\Gamma_F\longrightarrow X$ induces a birational morphism (as pointed out here this other assumption could be removed, feel free to share any thoughts).

If $X$ and $Y$ are disjoint union (possibly not finite) of irreducible projective varieties, say $X=\bigsqcup_{\alpha} X_\alpha$ and $Y=\sqcup_{\beta}Y_\beta$, than we say that $F$ is a CAM if for any $\alpha$ $F$ induces, by restriction, a continuos algebraic map $F\colon X_\alpha\longrightarrow Y_\beta$. This definition is that given by C. Peters and S. Kosarew.

In order to check that $\mathcal{C}_{p}(X,f)$ does not depend from $f$ we have to check that if $f'\colon X\subset \mathbb{P}^m$ then there is an algebraic homeomorphism (obvius meaning) from $\mathcal{C}_p(X,f*f')\longrightarrow \mathcal{C}_p(X,f)$, where $f\ast f'$ is the Segre embedding of $X\times X$ into $\mathbb{P}^{nm+n+m}$. It seems that the reference for this is Hoyt's paper On the Chow Bunches for Different Projective Embeddings of a Complete Variety. In this paper Hoyt shows that the map $\mathcal{C}_p(X,f*f')\longrightarrow \mathcal{C}_p(X,f)$ we are looking for is that which takes a cycle $(f\ast f')(\xi)$ to $f(\xi)$ which amounts to take the Chow form $F^{(f\ast f')(\xi)}(X_{qi}Y_{qj})=\sum_{n=1}^t A_n(X)M_n(Y)$ to $F^{f(\xi)}(X)=\mathsf{gdc}(A_1(X),\ldots,A_t(X))$.

In the Lemma 2 he proves that if $L$ is some projective space containing $(A_1,\ldots, A_t)$ and $L'$ is the space of forms of degree $d$, then the map the set $U\subset L$ of elements $(A_1,\ldots, A_t)$ having gdc of degree $d$ is locally closed and the map $g\colon U\longrightarrow L'$ taking $(A_1,\ldots,A_t)$ to $\mathsf{gdc}(A_1,\ldots, A_t)$ is a morphism. Then he writes:

enter image description here

where $Q(f\ast f')$ and $Q(f)$ are irreducible components of $\mathcal{C}_p(X,f\ast f')$ and $\mathcal{C}_p(X,f)$, rispectively.

Homestly I dont' understand how he uses the Lemma to get this conclusion. To be more precise, how can I conclude that the immage of $Q(f\ast f')$ via the gdc-map is contained in $Q(f)$?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.