# Pull-back of algebraic cycles under holomorphic maps

Let $f:X \to Y$ be a holomorphic map between two smooth complex projective manifolds. Is there a good notion of pull-back of algebraic cycles by $f$ which preserves degree in the following sense: Suppose $V_1, V_2$ be two cycles in $Y$ with the same Hilbert polynomial and suppose $f^{-1}(V_i)$ are algebraic varieties. Then will the pull-backs of $V_i$ by $f$ have the same dimension and degree (as algebraic cycles)? A small comparison with the divisor case is as follows: As far as I understand that pull-back of line bundles preserves degree in the sense as above. By identifying Picard group with the divisor class group we have a similar statement as above.

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Pull-back of line bundles, say on curves, does not preserve the degree. Instead, it multiplies the degree of the line bundles by the degree of the map. Is that what you want? The Hilbert polynomial depends on the projective embedding, specifically through the line bundle $\mathcal O(1)$. Do you want $F^* \mathcal O(1)=\mathcal O(1)$? Something else? – Will Sawin Mar 9 '13 at 23:09
@Sawin: I understand this fact. I meant that the ratio of the degrees of the pull-back of two line bundles with the same degree is $1$. That is what I try to explain in the beginning. – Naga Venkata Mar 9 '13 at 23:32

Yes, as long as $f^* \mathcal O(1)$ is a rational mutliple of $\mathcal O(1)$. However, we will preserve not the dimension, but the codimension, as suggested by the case of divisors. We use the standard notion of pullback of cycles in algebraic geometry (e.g. Hartshorne appendix $A$). This behaves well with respect to the intersection pairing, so $f^* A \cdot f^* B = f^*( A \cdot B)$.

Next recall that the degree of a cycle $V$ of dimension $d$ is just $V \cdot H^d$, where $H$ is the hyperplance class. Assume $H_Y = q f^* H_X$, and let $k$ be the relative dimension of $f$. Then the pullback of $V$ has dimension $k+d$, so

$f^* V \cdot H_X^{k+d} = f^* V_1 \cdot H_X^d \cdot H_X^k = q^d f^* V_1 \cdot f^* H_Y^d \cdot H_X^k= q^d f^* (V_1 \cdot H_Y^d) \cdot H_X^k$

but $V_1 \cdot H_Y^d$ is just the class of $n$ points, where $n$ is the degree of $V$, so the degree of $f^* V$ is the degree of $V$ times $q^d f^* (pt) \cdot H_X^k$.

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