In a tensor category, does $Xotimes Ycong 0$ imply $Ycong 0$ for non-zero $X$?
$begingroup$
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatorname{Hom}(1,1)cong k$.$^1$
Suppose we fix some non-zero object $X$ in such a category, and we take any old object $Y$. Can we conclude from $Xotimes Y cong 0$ that $Y$ has to be zero?
I know that in this setting the tensor product is (bi)exact. And I think the statement would be obvious if $otimes$ would reflect isomorphisms, but I don't feel it does.
Any hints?
$^1$ This question really only needs rigid abelian with bilinear $otimes$
category-theory abelian-categories monoidal-categories functors
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
$endgroup$
add a comment |
$begingroup$
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatorname{Hom}(1,1)cong k$.$^1$
Suppose we fix some non-zero object $X$ in such a category, and we take any old object $Y$. Can we conclude from $Xotimes Y cong 0$ that $Y$ has to be zero?
I know that in this setting the tensor product is (bi)exact. And I think the statement would be obvious if $otimes$ would reflect isomorphisms, but I don't feel it does.
Any hints?
$^1$ This question really only needs rigid abelian with bilinear $otimes$
category-theory abelian-categories monoidal-categories functors
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
$endgroup$
2
$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14
1
$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27
$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51
1
$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04
$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30
add a comment |
$begingroup$
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatorname{Hom}(1,1)cong k$.$^1$
Suppose we fix some non-zero object $X$ in such a category, and we take any old object $Y$. Can we conclude from $Xotimes Y cong 0$ that $Y$ has to be zero?
I know that in this setting the tensor product is (bi)exact. And I think the statement would be obvious if $otimes$ would reflect isomorphisms, but I don't feel it does.
Any hints?
$^1$ This question really only needs rigid abelian with bilinear $otimes$
category-theory abelian-categories monoidal-categories functors
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
$endgroup$
By a tensor category I mean a locally finite rigid $k$-linear abelian category with bilinear tensor product, and such that $operatorname{Hom}(1,1)cong k$.$^1$
Suppose we fix some non-zero object $X$ in such a category, and we take any old object $Y$. Can we conclude from $Xotimes Y cong 0$ that $Y$ has to be zero?
I know that in this setting the tensor product is (bi)exact. And I think the statement would be obvious if $otimes$ would reflect isomorphisms, but I don't feel it does.
Any hints?
$^1$ This question really only needs rigid abelian with bilinear $otimes$
category-theory abelian-categories monoidal-categories functors
category-theory abelian-categories monoidal-categories functors
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
edited Jul 26 '18 at 8:10
Jo Be
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
asked Jul 26 '18 at 7:54
Jo BeJo Be
6751519
6751519
2
$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14
1
$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27
$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51
1
$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04
$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30
add a comment |
2
$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14
1
$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27
$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51
1
$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04
$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30
2
2
$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14
$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14
1
1
$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27
$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27
$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51
$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51
1
1
$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04
$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04
$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30
$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The coevaluation map $1 rightarrow {}^*X otimes X$ is non-zero, so by simplicity of $1$ it's injective. But then by biexactness we have:
$$Y hookrightarrow {}^* X otimes X otimes Y cong 0.$$
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
$endgroup$
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
add a comment |
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%2f2863174%2fin-a-tensor-category-does-x-otimes-y-cong-0-imply-y-cong-0-for-non-zero-x%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
$begingroup$
The coevaluation map $1 rightarrow {}^*X otimes X$ is non-zero, so by simplicity of $1$ it's injective. But then by biexactness we have:
$$Y hookrightarrow {}^* X otimes X otimes Y cong 0.$$
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
$endgroup$
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
add a comment |
$begingroup$
The coevaluation map $1 rightarrow {}^*X otimes X$ is non-zero, so by simplicity of $1$ it's injective. But then by biexactness we have:
$$Y hookrightarrow {}^* X otimes X otimes Y cong 0.$$
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
$endgroup$
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
add a comment |
$begingroup$
The coevaluation map $1 rightarrow {}^*X otimes X$ is non-zero, so by simplicity of $1$ it's injective. But then by biexactness we have:
$$Y hookrightarrow {}^* X otimes X otimes Y cong 0.$$
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
$endgroup$
The coevaluation map $1 rightarrow {}^*X otimes X$ is non-zero, so by simplicity of $1$ it's injective. But then by biexactness we have:
$$Y hookrightarrow {}^* X otimes X otimes Y cong 0.$$
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
answered Jan 11 at 18:41
Noah SnyderNoah Snyder
7,65232955
7,65232955
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
add a comment |
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
$begingroup$
Thanks Noah. I think Qiaochu's commented was directed at my footnote. For your argument to work we obviously need simplicity of 1, which I dropped for some reason, I don't recall why - I usually only work in honest (finite) tensor cats.
$endgroup$
– Jo Be
Jan 11 at 18:47
add a comment |
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%2f2863174%2fin-a-tensor-category-does-x-otimes-y-cong-0-imply-y-cong-0-for-non-zero-x%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
$begingroup$
No, this is already false in $text{Vect} times text{Vect}$ with the pointwise tensor product.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 8:14
1
$begingroup$
$mathrm{Vect}$ is fd VS I assume, what do you mean by "pointwise tensor product" - $(V_1,V_2)otimes (W_1,W_2) equiv (V_1otimes W_1, V_2otimes W_2)$? Then the zero is $(0,0)$, and with $(V,0)$ and $(0,W)$, we get $(V,0)otimes (0,W)cong (0,0)$, right?
$endgroup$
– Jo Be
Jul 26 '18 at 8:27
$begingroup$
@QiaochuYuan: I forgot to ping, I hope it's alright I'm doing it now. Is the above the argument you had in mind?
$endgroup$
– Jo Be
Jul 26 '18 at 8:51
1
$begingroup$
Yes, that's right.
$endgroup$
– Qiaochu Yuan
Jul 26 '18 at 9:04
$begingroup$
@QiaochuYuan: Vect x Vect does not have End(1) = k!
$endgroup$
– Noah Snyder
Jan 11 at 18:30