Trying to understand morphisms between varieties
$begingroup$
Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.
(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)
I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.
ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$
algebraic-geometry
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
$endgroup$
add a comment |
$begingroup$
Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.
(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)
I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.
ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$
algebraic-geometry
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
$endgroup$
2
$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28
add a comment |
$begingroup$
Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.
(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)
I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.
ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$
algebraic-geometry
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
$endgroup$
Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.
(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)
I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.
ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$
algebraic-geometry
algebraic-geometry
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
asked Dec 22 '18 at 20:11
Johnny T.Johnny T.
5821415
5821415
2
$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28
add a comment |
2
$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28
2
2
$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28
$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
$endgroup$
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
1
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
1
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
1
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
|
show 2 more comments
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049809%2ftrying-to-understand-morphisms-between-varieties%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
$endgroup$
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
1
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
1
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
1
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
|
show 2 more comments
$begingroup$
You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
$endgroup$
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
1
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
1
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
1
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
|
show 2 more comments
$begingroup$
You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
$endgroup$
You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
edited Dec 22 '18 at 20:31
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
answered Dec 22 '18 at 20:24
LeventLevent
2,741925
2,741925
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
1
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
1
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
1
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
|
show 2 more comments
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
1
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
1
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
1
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34
1
1
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46
1
1
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51
1
1
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049809%2ftrying-to-understand-morphisms-between-varieties%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28