Exterior algebra as quotient algebra
This is my first time posting so I apologise for the stupid question. I am trying to understand the idea of exterior algebra as a quotient algebra. Let $T(V)$ be the tensor algebra of a vector space $V$. The exterior algebra is defined as
begin{equation}
Lambda(V)=T(V)/I,
end{equation}
where $I$ is the two-sided ideal generated by elements of the form $votimes v$ with $vin V$. This means the ideal is the set of elements
begin{equation}
I=left{sum_i r_iotimes votimes votimes s_ibigg|vin V,r_i,s_iin T(V) right}.
end{equation}
My understand of taking the quotient is that elements of the ideal should be identified as the additive identity of the algebra, $0_R$, under projection.
My question is as follows. Suppose we have the element
begin{equation}
aotimes b otimes cin V.
end{equation}
Now I add on the following element in the ideal $aotimes (-b+2a)otimes (-b+2a)in I$ -- this is in the ideal because it is of the form $aotimes votimes v$. Adding an element from the ideal should mean I remain in the same equivalence class, so
begin{equation}
aotimes b otimes csim aotimes botimes c+aotimes (-b+2a)otimes(-b+2a)=2aotimes2aotimes (c-b+2a).
end{equation}
The term on the right is in the form of an ideal. From this, I conclude that $aotimes botimes c$ is also in the ideal! This is blatantly wrong. It appears as if I can always add elements of the ideal to an element in $T(V)$ that is not in the ideal to obtain one that is in the ideal.
Where did I go wrong?
Thank you in advance!
differential-geometry exterior-algebra
asked Nov 25 '18 at 23:15
KinLong
11
add a comment |
This is my first time posting so I apologise for the stupid question. I am trying to understand the idea of exterior algebra as a quotient algebra. Let $T(V)$ be the tensor algebra of a vector space $V$. The exterior algebra is defined as
begin{equation}
Lambda(V)=T(V)/I,
end{equation}
where $I$ is the two-sided ideal generated by elements of the form $votimes v$ with $vin V$. This means the ideal is the set of elements
begin{equation}
I=left{sum_i r_iotimes votimes votimes s_ibigg|vin V,r_i,s_iin T(V) right}.
end{equation}
My understand of taking the quotient is that elements of the ideal should be identified as the additive identity of the algebra, $0_R$, under projection.
My question is as follows. Suppose we have the element
begin{equation}
aotimes b otimes cin V.
end{equation}
Now I add on the following element in the ideal $aotimes (-b+2a)otimes (-b+2a)in I$ -- this is in the ideal because it is of the form $aotimes votimes v$. Adding an element from the ideal should mean I remain in the same equivalence class, so
begin{equation}
aotimes b otimes csim aotimes botimes c+aotimes (-b+2a)otimes(-b+2a)=2aotimes2aotimes (c-b+2a).
end{equation}
The term on the right is in the form of an ideal. From this, I conclude that $aotimes botimes c$ is also in the ideal! This is blatantly wrong. It appears as if I can always add elements of the ideal to an element in $T(V)$ that is not in the ideal to obtain one that is in the ideal.
Where did I go wrong?
Thank you in advance!
differential-geometry exterior-algebra
asked Nov 25 '18 at 23:15
KinLong
11
That's not how addition of tensors works. $xotimes y +zotimes w neq (x+z)otimes (y+w)$ in general.
– Matthew Towers
Nov 25 '18 at 23:29
1
Oh, of course. Thank you! Can't believe I missed that!
– KinLong
Nov 25 '18 at 23:56
add a comment |
This is my first time posting so I apologise for the stupid question. I am trying to understand the idea of exterior algebra as a quotient algebra. Let $T(V)$ be the tensor algebra of a vector space $V$. The exterior algebra is defined as
begin{equation}
Lambda(V)=T(V)/I,
end{equation}
where $I$ is the two-sided ideal generated by elements of the form $votimes v$ with $vin V$. This means the ideal is the set of elements
begin{equation}
I=left{sum_i r_iotimes votimes votimes s_ibigg|vin V,r_i,s_iin T(V) right}.
end{equation}
My understand of taking the quotient is that elements of the ideal should be identified as the additive identity of the algebra, $0_R$, under projection.
My question is as follows. Suppose we have the element
begin{equation}
aotimes b otimes cin V.
end{equation}
Now I add on the following element in the ideal $aotimes (-b+2a)otimes (-b+2a)in I$ -- this is in the ideal because it is of the form $aotimes votimes v$. Adding an element from the ideal should mean I remain in the same equivalence class, so
begin{equation}
aotimes b otimes csim aotimes botimes c+aotimes (-b+2a)otimes(-b+2a)=2aotimes2aotimes (c-b+2a).
end{equation}
The term on the right is in the form of an ideal. From this, I conclude that $aotimes botimes c$ is also in the ideal! This is blatantly wrong. It appears as if I can always add elements of the ideal to an element in $T(V)$ that is not in the ideal to obtain one that is in the ideal.
Where did I go wrong?
Thank you in advance!
differential-geometry exterior-algebra
asked Nov 25 '18 at 23:15
KinLong
11
This is my first time posting so I apologise for the stupid question. I am trying to understand the idea of exterior algebra as a quotient algebra. Let $T(V)$ be the tensor algebra of a vector space $V$. The exterior algebra is defined as
begin{equation}
Lambda(V)=T(V)/I,
end{equation}
where $I$ is the two-sided ideal generated by elements of the form $votimes v$ with $vin V$. This means the ideal is the set of elements
begin{equation}
I=left{sum_i r_iotimes votimes votimes s_ibigg|vin V,r_i,s_iin T(V) right}.
end{equation}
My understand of taking the quotient is that elements of the ideal should be identified as the additive identity of the algebra, $0_R$, under projection.
My question is as follows. Suppose we have the element
begin{equation}
aotimes b otimes cin V.
end{equation}
Now I add on the following element in the ideal $aotimes (-b+2a)otimes (-b+2a)in I$ -- this is in the ideal because it is of the form $aotimes votimes v$. Adding an element from the ideal should mean I remain in the same equivalence class, so
begin{equation}
aotimes b otimes csim aotimes botimes c+aotimes (-b+2a)otimes(-b+2a)=2aotimes2aotimes (c-b+2a).
end{equation}
The term on the right is in the form of an ideal. From this, I conclude that $aotimes botimes c$ is also in the ideal! This is blatantly wrong. It appears as if I can always add elements of the ideal to an element in $T(V)$ that is not in the ideal to obtain one that is in the ideal.
Where did I go wrong?
Thank you in advance!
differential-geometry exterior-algebra
differential-geometry exterior-algebra
asked Nov 25 '18 at 23:15
KinLong
11
asked Nov 25 '18 at 23:15
KinLong
11
asked Nov 25 '18 at 23:15
KinLong
11
asked Nov 25 '18 at 23:15
KinLong
11
asked Nov 25 '18 at 23:15
KinLong
11
11
That's not how addition of tensors works. $xotimes y +zotimes w neq (x+z)otimes (y+w)$ in general.
– Matthew Towers
Nov 25 '18 at 23:29
1
Oh, of course. Thank you! Can't believe I missed that!
– KinLong
Nov 25 '18 at 23:56
add a comment |
That's not how addition of tensors works. $xotimes y +zotimes w neq (x+z)otimes (y+w)$ in general.
– Matthew Towers
Nov 25 '18 at 23:29
1
Oh, of course. Thank you! Can't believe I missed that!
– KinLong
Nov 25 '18 at 23:56
That's not how addition of tensors works. $xotimes y +zotimes w neq (x+z)otimes (y+w)$ in general.
– Matthew Towers
Nov 25 '18 at 23:29
That's not how addition of tensors works. $xotimes y +zotimes w neq (x+z)otimes (y+w)$ in general.
– Matthew Towers
Nov 25 '18 at 23:29
1
1
Oh, of course. Thank you! Can't believe I missed that!
– KinLong
Nov 25 '18 at 23:56
Oh, of course. Thank you! Can't believe I missed that!
– KinLong
Nov 25 '18 at 23:56
add a comment |
0
active
oldest
votes
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%2f3013554%2fexterior-algebra-as-quotient-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3013554%2fexterior-algebra-as-quotient-algebra%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
That's not how addition of tensors works. $xotimes y +zotimes w neq (x+z)otimes (y+w)$ in general.
– Matthew Towers
Nov 25 '18 at 23:29
1
Oh, of course. Thank you! Can't believe I missed that!
– KinLong
Nov 25 '18 at 23:56