Reflection Functor of a Quiver
$begingroup$
Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.
Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$
The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
end{cases}$$
Alternatively one can think of the sequence
$$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
in which the first map is the trivial map and the second is inclusion.
My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?
My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.
Is this correct?
If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?
Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?
As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.
representation-theory quiver
$endgroup$
add a comment |
$begingroup$
Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.
Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$
The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
end{cases}$$
Alternatively one can think of the sequence
$$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
in which the first map is the trivial map and the second is inclusion.
My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?
My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.
Is this correct?
If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?
Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?
As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.
representation-theory quiver
$endgroup$
add a comment |
$begingroup$
Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.
Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$
The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
end{cases}$$
Alternatively one can think of the sequence
$$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
in which the first map is the trivial map and the second is inclusion.
My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?
My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.
Is this correct?
If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?
Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?
As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.
representation-theory quiver
$endgroup$
Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.
Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$
The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
end{cases}$$
Alternatively one can think of the sequence
$$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
in which the first map is the trivial map and the second is inclusion.
My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?
My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.
Is this correct?
If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?
Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?
As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.
representation-theory quiver
representation-theory quiver
edited Dec 18 '18 at 0:57
user551642
asked Dec 18 '18 at 0:24
user551642user551642
2527
2527
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think I have figured it out;
let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
$$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$
Although I still don't know anything about the dimension in general.
$endgroup$
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
add a comment |
$begingroup$
If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.
$endgroup$
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3044639%2freflection-functor-of-a-quiver%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think I have figured it out;
let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
$$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$
Although I still don't know anything about the dimension in general.
$endgroup$
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
add a comment |
$begingroup$
I think I have figured it out;
let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
$$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$
Although I still don't know anything about the dimension in general.
$endgroup$
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
add a comment |
$begingroup$
I think I have figured it out;
let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
$$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$
Although I still don't know anything about the dimension in general.
$endgroup$
I think I have figured it out;
let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
$$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$
Although I still don't know anything about the dimension in general.
edited Dec 18 '18 at 1:38
answered Dec 18 '18 at 1:33
user551642user551642
2527
2527
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
add a comment |
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
$endgroup$
– Mike Pierce
Dec 18 '18 at 3:56
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
$begingroup$
Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
$endgroup$
– Mike Pierce
Dec 18 '18 at 4:03
add a comment |
$begingroup$
If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.
$endgroup$
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
add a comment |
$begingroup$
If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.
$endgroup$
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
add a comment |
$begingroup$
If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.
$endgroup$
If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.
edited Dec 18 '18 at 17:02
answered Dec 18 '18 at 16:28
StephenStephen
10.7k12439
10.7k12439
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
add a comment |
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
$begingroup$
Nice, this will be very useful. Thank you.
$endgroup$
– user551642
Dec 18 '18 at 23:29
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%2f3044639%2freflection-functor-of-a-quiver%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