Mixing
Fourier Mukai transform of skyscraper sheaf through flat kernel.
$begingroup$
Let $X,Y $ be smooth projective varieties over a field $k$.
The following example is taken from Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" [example 5.4.vi]:
"Suppose $mathcal{P}$ is a coherent sheaf on $Xtimes Y$ flat over $X$ and consider the Fourier Mukai Transform $Phi_mathcal{P}$. If $xin X$ is a closed point with $k(x)cong k$, then $Phi_mathcal{P}(k(x))cong P_{|xtimes Y}$ where $mathcal{P}_{|xtimes Y}$ is considered as a sheaf on $Y$ via the second projection."
Let $q: Xtimes Yto X$ and $p: Xtimes Yto Y$ denote the projections. Let $i: xtimes Yto Xtimes Y$ denote the closed immersion obtained from $xto X$ by base change.
I calculated $q^*k(x)cong i_*i^*mathcal{O}_{Xtimes Y}$ and then I got
$q^*k(x)otimesmathcal{P}cong i_*i^*mathcal{O}_{Xtimes Y}otimes mathcal{P}cong i_*(i^*mathcal{O}_{Xtimes Y}otimes i^*mathcal{P})cong i_*i^*mathcal{P}$ via the projection formula.
Applying the right derived functor $Rp_*$ yields $Rp_*i_*i^*mathcal{P}cong R(p_*i_*)(i^*mathcal{P})cong q'_*i^*mathcal{P}$ where $q':xtimes Yto Y$ is the second projection. This is the result from the example. I do not see where I have used flatness of $mathcal{P}$ over $X$ though. Where do I need it ? Thanks a lot .
EDIT: Flatness over $X$ should play a role in calculating the derived tensor. I actually assumed the derived tensor product equals the usual tensor product when calculating the FMT.
Is this implied by $mathcal{P}$ being flat over $X$?
algebraic-geometry
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
$endgroup$
add a comment |
$begingroup$
Let $X,Y $ be smooth projective varieties over a field $k$.
The following example is taken from Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" [example 5.4.vi]:
"Suppose $mathcal{P}$ is a coherent sheaf on $Xtimes Y$ flat over $X$ and consider the Fourier Mukai Transform $Phi_mathcal{P}$. If $xin X$ is a closed point with $k(x)cong k$, then $Phi_mathcal{P}(k(x))cong P_{|xtimes Y}$ where $mathcal{P}_{|xtimes Y}$ is considered as a sheaf on $Y$ via the second projection."
Let $q: Xtimes Yto X$ and $p: Xtimes Yto Y$ denote the projections. Let $i: xtimes Yto Xtimes Y$ denote the closed immersion obtained from $xto X$ by base change.
I calculated $q^*k(x)cong i_*i^*mathcal{O}_{Xtimes Y}$ and then I got
$q^*k(x)otimesmathcal{P}cong i_*i^*mathcal{O}_{Xtimes Y}otimes mathcal{P}cong i_*(i^*mathcal{O}_{Xtimes Y}otimes i^*mathcal{P})cong i_*i^*mathcal{P}$ via the projection formula.
Applying the right derived functor $Rp_*$ yields $Rp_*i_*i^*mathcal{P}cong R(p_*i_*)(i^*mathcal{P})cong q'_*i^*mathcal{P}$ where $q':xtimes Yto Y$ is the second projection. This is the result from the example. I do not see where I have used flatness of $mathcal{P}$ over $X$ though. Where do I need it ? Thanks a lot .
EDIT: Flatness over $X$ should play a role in calculating the derived tensor. I actually assumed the derived tensor product equals the usual tensor product when calculating the FMT.
Is this implied by $mathcal{P}$ being flat over $X$?
algebraic-geometry
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
$endgroup$
$begingroup$
Did you figure this out? $i^* O_{X times Y} = O_{x times Y}$, and tensoring against this (derived or not) is the identity, I think...
$endgroup$
– Lorenzo
Jan 29 '17 at 21:47
add a comment |
$begingroup$
Let $X,Y $ be smooth projective varieties over a field $k$.
The following example is taken from Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" [example 5.4.vi]:
"Suppose $mathcal{P}$ is a coherent sheaf on $Xtimes Y$ flat over $X$ and consider the Fourier Mukai Transform $Phi_mathcal{P}$. If $xin X$ is a closed point with $k(x)cong k$, then $Phi_mathcal{P}(k(x))cong P_{|xtimes Y}$ where $mathcal{P}_{|xtimes Y}$ is considered as a sheaf on $Y$ via the second projection."
Let $q: Xtimes Yto X$ and $p: Xtimes Yto Y$ denote the projections. Let $i: xtimes Yto Xtimes Y$ denote the closed immersion obtained from $xto X$ by base change.
I calculated $q^*k(x)cong i_*i^*mathcal{O}_{Xtimes Y}$ and then I got
$q^*k(x)otimesmathcal{P}cong i_*i^*mathcal{O}_{Xtimes Y}otimes mathcal{P}cong i_*(i^*mathcal{O}_{Xtimes Y}otimes i^*mathcal{P})cong i_*i^*mathcal{P}$ via the projection formula.
Applying the right derived functor $Rp_*$ yields $Rp_*i_*i^*mathcal{P}cong R(p_*i_*)(i^*mathcal{P})cong q'_*i^*mathcal{P}$ where $q':xtimes Yto Y$ is the second projection. This is the result from the example. I do not see where I have used flatness of $mathcal{P}$ over $X$ though. Where do I need it ? Thanks a lot .
EDIT: Flatness over $X$ should play a role in calculating the derived tensor. I actually assumed the derived tensor product equals the usual tensor product when calculating the FMT.
Is this implied by $mathcal{P}$ being flat over $X$?
algebraic-geometry
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
$endgroup$
Let $X,Y $ be smooth projective varieties over a field $k$.
The following example is taken from Huybrecht's "Fourier Mukai Transforms in Algebraic Geometry" [example 5.4.vi]:
"Suppose $mathcal{P}$ is a coherent sheaf on $Xtimes Y$ flat over $X$ and consider the Fourier Mukai Transform $Phi_mathcal{P}$. If $xin X$ is a closed point with $k(x)cong k$, then $Phi_mathcal{P}(k(x))cong P_{|xtimes Y}$ where $mathcal{P}_{|xtimes Y}$ is considered as a sheaf on $Y$ via the second projection."
Let $q: Xtimes Yto X$ and $p: Xtimes Yto Y$ denote the projections. Let $i: xtimes Yto Xtimes Y$ denote the closed immersion obtained from $xto X$ by base change.
I calculated $q^*k(x)cong i_*i^*mathcal{O}_{Xtimes Y}$ and then I got
$q^*k(x)otimesmathcal{P}cong i_*i^*mathcal{O}_{Xtimes Y}otimes mathcal{P}cong i_*(i^*mathcal{O}_{Xtimes Y}otimes i^*mathcal{P})cong i_*i^*mathcal{P}$ via the projection formula.
Applying the right derived functor $Rp_*$ yields $Rp_*i_*i^*mathcal{P}cong R(p_*i_*)(i^*mathcal{P})cong q'_*i^*mathcal{P}$ where $q':xtimes Yto Y$ is the second projection. This is the result from the example. I do not see where I have used flatness of $mathcal{P}$ over $X$ though. Where do I need it ? Thanks a lot .
EDIT: Flatness over $X$ should play a role in calculating the derived tensor. I actually assumed the derived tensor product equals the usual tensor product when calculating the FMT.
Is this implied by $mathcal{P}$ being flat over $X$?
algebraic-geometry
algebraic-geometry
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
edited Feb 1 '12 at 12:06
Carsten
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
asked Feb 1 '12 at 9:35
CarstenCarsten
384111
384111
$begingroup$
Did you figure this out? $i^* O_{X times Y} = O_{x times Y}$, and tensoring against this (derived or not) is the identity, I think...
$endgroup$
– Lorenzo
Jan 29 '17 at 21:47
add a comment |
$begingroup$
Did you figure this out? $i^* O_{X times Y} = O_{x times Y}$, and tensoring against this (derived or not) is the identity, I think...
$endgroup$
– Lorenzo
Jan 29 '17 at 21:47
$begingroup$
Did you figure this out? $i^* O_{X times Y} = O_{x times Y}$, and tensoring against this (derived or not) is the identity, I think...
$endgroup$
– Lorenzo
Jan 29 '17 at 21:47
$begingroup$
Did you figure this out? $i^* O_{X times Y} = O_{x times Y}$, and tensoring against this (derived or not) is the identity, I think...
$endgroup$
– Lorenzo
Jan 29 '17 at 21:47
add a comment |
1 Answer
1
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oldest
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$begingroup$
I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.
The final answer you get is $q’_* i^* mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $mathcal{P}$ is flat over $X$ we know that the functor $mathcal{G} mapsto q^*mathcal{G} otimes mathcal{P}$ from modules over $X$ to modules over $X times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) otimes mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get
$$
q^* k(x) otimes mathcal{P} simeq i_* i^* mathcal{P}
$$
as sheaves. We now conclude applying the functor $Rp_*$, which yields
$$
Phi_{mathcal{P}} (k(x)) simeq q’_* (i^* mathcal{P})
$$
as a sheaf.
answered Jan 21 at 16:37
FedericoFederico
918313
$endgroup$
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
add a comment |
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$begingroup$
I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.
The final answer you get is $q’_* i^* mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $mathcal{P}$ is flat over $X$ we know that the functor $mathcal{G} mapsto q^*mathcal{G} otimes mathcal{P}$ from modules over $X$ to modules over $X times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) otimes mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get
$$
q^* k(x) otimes mathcal{P} simeq i_* i^* mathcal{P}
$$
as sheaves. We now conclude applying the functor $Rp_*$, which yields
$$
Phi_{mathcal{P}} (k(x)) simeq q’_* (i^* mathcal{P})
$$
as a sheaf.
answered Jan 21 at 16:37
FedericoFederico
918313
$endgroup$
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
add a comment |
$begingroup$
I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.
The final answer you get is $q’_* i^* mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $mathcal{P}$ is flat over $X$ we know that the functor $mathcal{G} mapsto q^*mathcal{G} otimes mathcal{P}$ from modules over $X$ to modules over $X times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) otimes mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get
$$
q^* k(x) otimes mathcal{P} simeq i_* i^* mathcal{P}
$$
as sheaves. We now conclude applying the functor $Rp_*$, which yields
$$
Phi_{mathcal{P}} (k(x)) simeq q’_* (i^* mathcal{P})
$$
as a sheaf.
answered Jan 21 at 16:37
FedericoFederico
918313
$endgroup$
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
add a comment |
$begingroup$
I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.
The final answer you get is $q’_* i^* mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $mathcal{P}$ is flat over $X$ we know that the functor $mathcal{G} mapsto q^*mathcal{G} otimes mathcal{P}$ from modules over $X$ to modules over $X times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) otimes mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get
$$
q^* k(x) otimes mathcal{P} simeq i_* i^* mathcal{P}
$$
as sheaves. We now conclude applying the functor $Rp_*$, which yields
$$
Phi_{mathcal{P}} (k(x)) simeq q’_* (i^* mathcal{P})
$$
as a sheaf.
answered Jan 21 at 16:37
FedericoFederico
918313
$endgroup$
I know it’s a very old question, but I think I figured out what’s the solution, so I post the answer.
The final answer you get is $q’_* i^* mathcal{P} $, however, you don’t know whether this is a sheaf or a true complex. First we notice that $q’_*$ need not be derived being an isomorphism, as the closed point was chosen such that $k(x) simeq k$ (I’m quite sure about this, as it works affine locally, but correct me if I’m wrong). As $mathcal{P}$ is flat over $X$ we know that the functor $mathcal{G} mapsto q^*mathcal{G} otimes mathcal{P}$ from modules over $X$ to modules over $X times Y$ is exact, where $q$ is the projection. Therefore, $q^* k(x) otimes mathcal{P}$ is a sheaf and the tensor product is not derived. Hence, we can use the stand projection formula (not the derived one) as $i_*$ is also exact, and therefore not derived, to get
$$
q^* k(x) otimes mathcal{P} simeq i_* i^* mathcal{P}
$$
as sheaves. We now conclude applying the functor $Rp_*$, which yields
$$
Phi_{mathcal{P}} (k(x)) simeq q’_* (i^* mathcal{P})
$$
as a sheaf.
answered Jan 21 at 16:37
FedericoFederico
918313
edited Jan 23 at 21:00
answered Jan 21 at 16:37
FedericoFederico
918313
answered Jan 21 at 16:37
FedericoFederico
918313
answered Jan 21 at 16:37
FedericoFederico
918313
918313
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
add a comment |
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Thanks for the answer. Looks fine, though I my derived categories are a bit rusty. I wonder: where is the change of hypothesis?
$endgroup$
– Carsten
Jan 23 at 20:31
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
$begingroup$
Sorry, that was a part of the previous post I wrote. I figured out the first answer was wrong and I edited the post, I forgot to delete that part. There's no change of hypothesis.
$endgroup$
– Federico
Jan 23 at 21:00
add a comment |
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$begingroup$
Did you figure this out? $i^* O_{X times Y} = O_{x times Y}$, and tensoring against this (derived or not) is the identity, I think...
$endgroup$
– Lorenzo
Jan 29 '17 at 21:47