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Let $X$ be a smooth projective variety, $Z$ a smooth subvariety of $X$, and let $f:\widetilde{X}\to X$ be the blow-up of $X$ along $Z$. Then for a subvariety $V\subset X$, we have two cohomology classes on $\widetilde{X}$: The pullback $f^*[V]$ and $[\widetilde{V}]$, the class of the proper transform of $V$. My question is the following:

Assuming $V\not\subseteq Z$ (but not that the intersection $V\cap Z$ is proper!), is the class $f^*[V]$ effective (i.e., a non-negative sum of classes of subvarieties)?

If the intersection $V\cap Z$ is transversal, then $f^*[V]=[\widetilde V]$. More generally, Fulton's book (Chapter 6) gives a formula relating $f^*[V]$, $[\widetilde{V}]$ and the Segre classes of the intersection $V\cap Z$, but it is not clear (to me at least) whether this gives an effective expression for $f^*[V]$.

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    $\begingroup$ I actually suspect the answer is "no", and my idea would be to create an example that is a toric variety (so that it is easy to say which cycles are effective). The OP does not say that $V$ and $Z$ even intersect dimensionally transversally. The OP even allows that $V$ might be a proper subvariety of $Z$. $\endgroup$ – Jason Starr Jul 17 '17 at 9:24
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    $\begingroup$ I agree. When we allow $V\subset Z$, I believe that $Z=P^2$ embedded as the zero-section in the total space of $\Omega_{P^2}$ could be a potential counterexample. $\endgroup$ – Jonathan Frink Jul 17 '17 at 17:13
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I guess you are right... :)
It seems to me that this indeed fails quite often.

Let $d=\mathrm{codim}_XZ$ and $P=f^{-1}Z\subseteq \widetilde X$ the exceptional divisor. By the assumptions $P\to Z$ is a $\mathbb P^{d-1}$-bundle.

First note that as long as $d\leq 2$, then the statement is probably true for the simple reason that if $V\neq X$, then $\dim (V\cap Z)\leq \dim V-1$ and hence the largest dimensional part of the pull-back of the Segre class $s(V\cap Z, V)$ has dimension at most $\dim V$ and hence that's the only component that makes a meaningful appearance in the formula for $f^*[V]$.

So, let's go to $d=3$. In this case $P\to Z$ is a $\mathbb P^{2}$-bundle. Now let $V\subseteq X$ be a surface such that $C=V\cap Z\subseteq V$ is a Cartier divisor in $V$. Then the Segre class in question is $$ s(C,V)=[C]-[C]^2. $$ Since we have a nice blow up, the "mysterious" excess normal bundle is just the relative tangent bundle $T_{P/Z}$, but this actually doesn't really matter, all we need is that there is a short exact sequence, $$ 0\to \mathscr O_P(-1) \to f^*N \to E\to 0, $$ where $N=N_{Z/X}$ is the normal bundle of $Z$ in $X$.

So we need the class $\left(c(E)\cdot_P f^*([C]-[C]^2)\right)_2$.

Observe that $f^*[C]$ has dimension $3$ and $f^*[C]^2=f^*([C]\cdot_V[C])$ has dimension $2$, so we need to intersect the first with $c_1(E)=c_1(f^*N)-c_1(\mathscr O_P(-1) )$ and the second with $c_0(E)=1$. Using that $$ c_1(f^*N)\cdot_P f^*[C]= f^*(c_1(N)\cdot_Z [C]), $$ we get that $$ \left(c(E)\cdot_P f^*([C]-[C]^2)\right)_2= f^*(c_1(N)\cdot_Z [C]-[C]^2)+ c_1(\mathscr O_P(1))\cdot_P f^*[C]. $$ So, it seems that if the intersection $C=V\cap Z$ is such that

  1. $C^2>0$ on $V$, and
  2. $c_1(N)\cdot_Z C<0$ on $Z$,

then the above class will not be effective. Note that these choices make the $f^*$ part of the class negative effective while the other class is "horizontal", so it is unlikely to cancel the "vertical" classes, but in any case one can probably play around with $V$ and $Z$ to make this more precise.

It seems that if the codimension of $Z$ is even higher and the intersection $V\cap Z$ is still large in $V$, then there is even more that can go wrong.

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