Mixing
What is $ v_1wedge v_2 wedge cdots wedge v_k$?
$begingroup$
If $T in wedge^k(V)$ and $S in wedge^l(V)$, then the definition of their wedge product is
$$T wedge S := frac{(k+l)!}{ k ! l!}text{Alt}(T otimes S) in wedge^{k+l}(V).$$
Meanwhile, we have that
$$text{Alt}(T) = frac{1}{k!}sum_{pi in S_p} {operatorname{sgn} , (pi)} T^pi.$$
Now my professor defines for $v_i in V$
$$v_1wedge v_2 wedge cdots wedge v_k := mathrm{Alt} (v_1 otimes v_2 ...otimes v_k) .$$
My question: do we need to define $v_1wedge v_2 wedge cdots wedge v_k$? I think we can derive it by two preceding definitions, but unfortunately I'm getting
$ k! , operatorname{Alt} (v_1 otimes v_2 ...otimes v_k)$. Is that $k!$ I'm getting a miscalculation?
tensor-products tensors multilinear-algebra exterior-algebra
asked Jul 30 '16 at 15:11
badaanbadaan
263
$endgroup$
add a comment |
$begingroup$
If $T in wedge^k(V)$ and $S in wedge^l(V)$, then the definition of their wedge product is
$$T wedge S := frac{(k+l)!}{ k ! l!}text{Alt}(T otimes S) in wedge^{k+l}(V).$$
Meanwhile, we have that
$$text{Alt}(T) = frac{1}{k!}sum_{pi in S_p} {operatorname{sgn} , (pi)} T^pi.$$
Now my professor defines for $v_i in V$
$$v_1wedge v_2 wedge cdots wedge v_k := mathrm{Alt} (v_1 otimes v_2 ...otimes v_k) .$$
My question: do we need to define $v_1wedge v_2 wedge cdots wedge v_k$? I think we can derive it by two preceding definitions, but unfortunately I'm getting
$ k! , operatorname{Alt} (v_1 otimes v_2 ...otimes v_k)$. Is that $k!$ I'm getting a miscalculation?
tensor-products tensors multilinear-algebra exterior-algebra
asked Jul 30 '16 at 15:11
badaanbadaan
263
$endgroup$
1
$begingroup$
I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+ell)!/k!ell!$ is designed specifically to overcome this problem.
$endgroup$
– Aaron
Jul 30 '16 at 17:51
1
$begingroup$
Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy.
$endgroup$
– Qiaochu Yuan
Jul 31 '16 at 20:52
add a comment |
$begingroup$
If $T in wedge^k(V)$ and $S in wedge^l(V)$, then the definition of their wedge product is
$$T wedge S := frac{(k+l)!}{ k ! l!}text{Alt}(T otimes S) in wedge^{k+l}(V).$$
Meanwhile, we have that
$$text{Alt}(T) = frac{1}{k!}sum_{pi in S_p} {operatorname{sgn} , (pi)} T^pi.$$
Now my professor defines for $v_i in V$
$$v_1wedge v_2 wedge cdots wedge v_k := mathrm{Alt} (v_1 otimes v_2 ...otimes v_k) .$$
My question: do we need to define $v_1wedge v_2 wedge cdots wedge v_k$? I think we can derive it by two preceding definitions, but unfortunately I'm getting
$ k! , operatorname{Alt} (v_1 otimes v_2 ...otimes v_k)$. Is that $k!$ I'm getting a miscalculation?
tensor-products tensors multilinear-algebra exterior-algebra
asked Jul 30 '16 at 15:11
badaanbadaan
263
$endgroup$
If $T in wedge^k(V)$ and $S in wedge^l(V)$, then the definition of their wedge product is
$$T wedge S := frac{(k+l)!}{ k ! l!}text{Alt}(T otimes S) in wedge^{k+l}(V).$$
Meanwhile, we have that
$$text{Alt}(T) = frac{1}{k!}sum_{pi in S_p} {operatorname{sgn} , (pi)} T^pi.$$
Now my professor defines for $v_i in V$
$$v_1wedge v_2 wedge cdots wedge v_k := mathrm{Alt} (v_1 otimes v_2 ...otimes v_k) .$$
My question: do we need to define $v_1wedge v_2 wedge cdots wedge v_k$? I think we can derive it by two preceding definitions, but unfortunately I'm getting
$ k! , operatorname{Alt} (v_1 otimes v_2 ...otimes v_k)$. Is that $k!$ I'm getting a miscalculation?
tensor-products tensors multilinear-algebra exterior-algebra
tensor-products tensors multilinear-algebra exterior-algebra
asked Jul 30 '16 at 15:11
badaanbadaan
263
asked Jul 30 '16 at 15:11
badaanbadaan
263
edited Aug 3 '16 at 9:53
badaan
asked Jul 30 '16 at 15:11
badaanbadaan
263
asked Jul 30 '16 at 15:11
badaanbadaan
263
asked Jul 30 '16 at 15:11
badaanbadaan
263
263
1
$begingroup$
I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+ell)!/k!ell!$ is designed specifically to overcome this problem.
$endgroup$
– Aaron
Jul 30 '16 at 17:51
1
$begingroup$
Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy.
$endgroup$
– Qiaochu Yuan
Jul 31 '16 at 20:52
add a comment |
1
$begingroup$
I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+ell)!/k!ell!$ is designed specifically to overcome this problem.
$endgroup$
– Aaron
Jul 30 '16 at 17:51
1
$begingroup$
Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy.
$endgroup$
– Qiaochu Yuan
Jul 31 '16 at 20:52
1
1
$begingroup$
I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+ell)!/k!ell!$ is designed specifically to overcome this problem.
$endgroup$
– Aaron
Jul 30 '16 at 17:51
$begingroup$
I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+ell)!/k!ell!$ is designed specifically to overcome this problem.
$endgroup$
– Aaron
Jul 30 '16 at 17:51
1
1
$begingroup$
Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy.
$endgroup$
– Qiaochu Yuan
Jul 31 '16 at 20:52
$begingroup$
Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy.
$endgroup$
– Qiaochu Yuan
Jul 31 '16 at 20:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have not miscalculated. It's just that the exterior product of alternating tensors is defined in two different ways and you have been mixing them without noticing. Let me explain.
One definition of $wedge$ is:
given $omegainbigwedge^r{(mathbb{R}^n)^*}, etainbigwedge^s{(mathbb{R}^n)^*}$ two alternating tensors and given $e^1,...,e^n$ a basis of $mathbb{R}^n$,
$$
omegawedgeeta = text{Alt}(omegaotimeseta) = frac{1}{(r+s)!}sum_{sigmainmathcal{S}_{r+s}}{text{sgn}(sigma)spacesigma(omega_{i_1,...,i_r}eta_{i_{r+1}...,i_{r+s}}e^1otimesldotsotimes e^{r+s})} = frac{omega_{i_1,...,i_r}eta_{i_{r+1},...,i_{r+s}}}{(r+s)!}e^1wedgeldotswedge e^{r+s}
$$
With this definition, the exterior product (of alt. tensors) of two vectors coincides with the "exterior product of vectors", i.e. the tensor you get by doing $v_1$ wedged with $v_2$ is precisely $v_1wedge v_2$.
The determinant as an $n$-form turns out to be $n!space e^1wedgeldotswedge e^n$. This is not very convenient when dealing with differential forms. So, there's this other definition (which I've called $wedge'$ to distinguish it from the other):
$$
omegawedge'eta = frac{omega_{i_1,...,i_r}}{r!}frac{eta_{i_{r+1},...,i_{r+s}}}{s!}e^1wedgeldotswedge e^{r+s}
$$
You can think of this definition as the one you'd get had you applied the $text{Alt}$ operator before wedging and not after wedging. We notice that
$$
omegawedge'eta = frac{(r+s)!}{r!s!}omegawedgeeta = frac{(r+s)!}{r!s!}text{Alt}(omegaotimeseta)
$$
The apparent miscalculation you were getting is just the result of using this second definition, which is not compatible with the exterior product of vectors.
In fact, what you've found is that
$$
e^1wedge'ldotswedge'e^n = n! space e^1 wedgeldotswedge e^n = text{det}
$$
and, if I didn't misunderstand this when I learnt it, $e^i = text{d}x^i$, which makes the element of volume $text{d}x^1 ldots text{d}x^n$ the determinant (as an $n$-form).
EDIT: as Qiaochu Yuan pointed out, these two definitions come from considering the set of alternating tensors either a quotient or a subspace of the set of tensors you're considering. Seems like I came a bit late to answer this, almost two years, but this could be useful for somebody else, so it hasn't been a waste of time to do this, anyway.
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
$endgroup$
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%2f1876045%2fwhat-is-v-1-wedge-v-2-wedge-cdots-wedge-v-k%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 have not miscalculated. It's just that the exterior product of alternating tensors is defined in two different ways and you have been mixing them without noticing. Let me explain.
One definition of $wedge$ is:
given $omegainbigwedge^r{(mathbb{R}^n)^*}, etainbigwedge^s{(mathbb{R}^n)^*}$ two alternating tensors and given $e^1,...,e^n$ a basis of $mathbb{R}^n$,
$$
omegawedgeeta = text{Alt}(omegaotimeseta) = frac{1}{(r+s)!}sum_{sigmainmathcal{S}_{r+s}}{text{sgn}(sigma)spacesigma(omega_{i_1,...,i_r}eta_{i_{r+1}...,i_{r+s}}e^1otimesldotsotimes e^{r+s})} = frac{omega_{i_1,...,i_r}eta_{i_{r+1},...,i_{r+s}}}{(r+s)!}e^1wedgeldotswedge e^{r+s}
$$
With this definition, the exterior product (of alt. tensors) of two vectors coincides with the "exterior product of vectors", i.e. the tensor you get by doing $v_1$ wedged with $v_2$ is precisely $v_1wedge v_2$.
The determinant as an $n$-form turns out to be $n!space e^1wedgeldotswedge e^n$. This is not very convenient when dealing with differential forms. So, there's this other definition (which I've called $wedge'$ to distinguish it from the other):
$$
omegawedge'eta = frac{omega_{i_1,...,i_r}}{r!}frac{eta_{i_{r+1},...,i_{r+s}}}{s!}e^1wedgeldotswedge e^{r+s}
$$
You can think of this definition as the one you'd get had you applied the $text{Alt}$ operator before wedging and not after wedging. We notice that
$$
omegawedge'eta = frac{(r+s)!}{r!s!}omegawedgeeta = frac{(r+s)!}{r!s!}text{Alt}(omegaotimeseta)
$$
The apparent miscalculation you were getting is just the result of using this second definition, which is not compatible with the exterior product of vectors.
In fact, what you've found is that
$$
e^1wedge'ldotswedge'e^n = n! space e^1 wedgeldotswedge e^n = text{det}
$$
and, if I didn't misunderstand this when I learnt it, $e^i = text{d}x^i$, which makes the element of volume $text{d}x^1 ldots text{d}x^n$ the determinant (as an $n$-form).
EDIT: as Qiaochu Yuan pointed out, these two definitions come from considering the set of alternating tensors either a quotient or a subspace of the set of tensors you're considering. Seems like I came a bit late to answer this, almost two years, but this could be useful for somebody else, so it hasn't been a waste of time to do this, anyway.
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
$endgroup$
add a comment |
$begingroup$
You have not miscalculated. It's just that the exterior product of alternating tensors is defined in two different ways and you have been mixing them without noticing. Let me explain.
One definition of $wedge$ is:
given $omegainbigwedge^r{(mathbb{R}^n)^*}, etainbigwedge^s{(mathbb{R}^n)^*}$ two alternating tensors and given $e^1,...,e^n$ a basis of $mathbb{R}^n$,
$$
omegawedgeeta = text{Alt}(omegaotimeseta) = frac{1}{(r+s)!}sum_{sigmainmathcal{S}_{r+s}}{text{sgn}(sigma)spacesigma(omega_{i_1,...,i_r}eta_{i_{r+1}...,i_{r+s}}e^1otimesldotsotimes e^{r+s})} = frac{omega_{i_1,...,i_r}eta_{i_{r+1},...,i_{r+s}}}{(r+s)!}e^1wedgeldotswedge e^{r+s}
$$
With this definition, the exterior product (of alt. tensors) of two vectors coincides with the "exterior product of vectors", i.e. the tensor you get by doing $v_1$ wedged with $v_2$ is precisely $v_1wedge v_2$.
The determinant as an $n$-form turns out to be $n!space e^1wedgeldotswedge e^n$. This is not very convenient when dealing with differential forms. So, there's this other definition (which I've called $wedge'$ to distinguish it from the other):
$$
omegawedge'eta = frac{omega_{i_1,...,i_r}}{r!}frac{eta_{i_{r+1},...,i_{r+s}}}{s!}e^1wedgeldotswedge e^{r+s}
$$
You can think of this definition as the one you'd get had you applied the $text{Alt}$ operator before wedging and not after wedging. We notice that
$$
omegawedge'eta = frac{(r+s)!}{r!s!}omegawedgeeta = frac{(r+s)!}{r!s!}text{Alt}(omegaotimeseta)
$$
The apparent miscalculation you were getting is just the result of using this second definition, which is not compatible with the exterior product of vectors.
In fact, what you've found is that
$$
e^1wedge'ldotswedge'e^n = n! space e^1 wedgeldotswedge e^n = text{det}
$$
and, if I didn't misunderstand this when I learnt it, $e^i = text{d}x^i$, which makes the element of volume $text{d}x^1 ldots text{d}x^n$ the determinant (as an $n$-form).
EDIT: as Qiaochu Yuan pointed out, these two definitions come from considering the set of alternating tensors either a quotient or a subspace of the set of tensors you're considering. Seems like I came a bit late to answer this, almost two years, but this could be useful for somebody else, so it hasn't been a waste of time to do this, anyway.
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
$endgroup$
add a comment |
$begingroup$
You have not miscalculated. It's just that the exterior product of alternating tensors is defined in two different ways and you have been mixing them without noticing. Let me explain.
One definition of $wedge$ is:
given $omegainbigwedge^r{(mathbb{R}^n)^*}, etainbigwedge^s{(mathbb{R}^n)^*}$ two alternating tensors and given $e^1,...,e^n$ a basis of $mathbb{R}^n$,
$$
omegawedgeeta = text{Alt}(omegaotimeseta) = frac{1}{(r+s)!}sum_{sigmainmathcal{S}_{r+s}}{text{sgn}(sigma)spacesigma(omega_{i_1,...,i_r}eta_{i_{r+1}...,i_{r+s}}e^1otimesldotsotimes e^{r+s})} = frac{omega_{i_1,...,i_r}eta_{i_{r+1},...,i_{r+s}}}{(r+s)!}e^1wedgeldotswedge e^{r+s}
$$
With this definition, the exterior product (of alt. tensors) of two vectors coincides with the "exterior product of vectors", i.e. the tensor you get by doing $v_1$ wedged with $v_2$ is precisely $v_1wedge v_2$.
The determinant as an $n$-form turns out to be $n!space e^1wedgeldotswedge e^n$. This is not very convenient when dealing with differential forms. So, there's this other definition (which I've called $wedge'$ to distinguish it from the other):
$$
omegawedge'eta = frac{omega_{i_1,...,i_r}}{r!}frac{eta_{i_{r+1},...,i_{r+s}}}{s!}e^1wedgeldotswedge e^{r+s}
$$
You can think of this definition as the one you'd get had you applied the $text{Alt}$ operator before wedging and not after wedging. We notice that
$$
omegawedge'eta = frac{(r+s)!}{r!s!}omegawedgeeta = frac{(r+s)!}{r!s!}text{Alt}(omegaotimeseta)
$$
The apparent miscalculation you were getting is just the result of using this second definition, which is not compatible with the exterior product of vectors.
In fact, what you've found is that
$$
e^1wedge'ldotswedge'e^n = n! space e^1 wedgeldotswedge e^n = text{det}
$$
and, if I didn't misunderstand this when I learnt it, $e^i = text{d}x^i$, which makes the element of volume $text{d}x^1 ldots text{d}x^n$ the determinant (as an $n$-form).
EDIT: as Qiaochu Yuan pointed out, these two definitions come from considering the set of alternating tensors either a quotient or a subspace of the set of tensors you're considering. Seems like I came a bit late to answer this, almost two years, but this could be useful for somebody else, so it hasn't been a waste of time to do this, anyway.
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
$endgroup$
You have not miscalculated. It's just that the exterior product of alternating tensors is defined in two different ways and you have been mixing them without noticing. Let me explain.
One definition of $wedge$ is:
given $omegainbigwedge^r{(mathbb{R}^n)^*}, etainbigwedge^s{(mathbb{R}^n)^*}$ two alternating tensors and given $e^1,...,e^n$ a basis of $mathbb{R}^n$,
$$
omegawedgeeta = text{Alt}(omegaotimeseta) = frac{1}{(r+s)!}sum_{sigmainmathcal{S}_{r+s}}{text{sgn}(sigma)spacesigma(omega_{i_1,...,i_r}eta_{i_{r+1}...,i_{r+s}}e^1otimesldotsotimes e^{r+s})} = frac{omega_{i_1,...,i_r}eta_{i_{r+1},...,i_{r+s}}}{(r+s)!}e^1wedgeldotswedge e^{r+s}
$$
With this definition, the exterior product (of alt. tensors) of two vectors coincides with the "exterior product of vectors", i.e. the tensor you get by doing $v_1$ wedged with $v_2$ is precisely $v_1wedge v_2$.
The determinant as an $n$-form turns out to be $n!space e^1wedgeldotswedge e^n$. This is not very convenient when dealing with differential forms. So, there's this other definition (which I've called $wedge'$ to distinguish it from the other):
$$
omegawedge'eta = frac{omega_{i_1,...,i_r}}{r!}frac{eta_{i_{r+1},...,i_{r+s}}}{s!}e^1wedgeldotswedge e^{r+s}
$$
You can think of this definition as the one you'd get had you applied the $text{Alt}$ operator before wedging and not after wedging. We notice that
$$
omegawedge'eta = frac{(r+s)!}{r!s!}omegawedgeeta = frac{(r+s)!}{r!s!}text{Alt}(omegaotimeseta)
$$
The apparent miscalculation you were getting is just the result of using this second definition, which is not compatible with the exterior product of vectors.
In fact, what you've found is that
$$
e^1wedge'ldotswedge'e^n = n! space e^1 wedgeldotswedge e^n = text{det}
$$
and, if I didn't misunderstand this when I learnt it, $e^i = text{d}x^i$, which makes the element of volume $text{d}x^1 ldots text{d}x^n$ the determinant (as an $n$-form).
EDIT: as Qiaochu Yuan pointed out, these two definitions come from considering the set of alternating tensors either a quotient or a subspace of the set of tensors you're considering. Seems like I came a bit late to answer this, almost two years, but this could be useful for somebody else, so it hasn't been a waste of time to do this, anyway.
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
edited Jan 22 at 17:14
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
answered May 26 '18 at 17:40
TeicDaunTeicDaun
66116
66116
add a comment |
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%2f1876045%2fwhat-is-v-1-wedge-v-2-wedge-cdots-wedge-v-k%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
1
$begingroup$
I would say that you have a miscalculation somewhere. A sensible definition of wedge product will be associative, and if you're getting a constant out in front like that, then it isn't associative. I imagine that the $(k+ell)!/k!ell!$ is designed specifically to overcome this problem.
$endgroup$
– Aaron
Jul 30 '16 at 17:51
1
$begingroup$
Your two definitions are clearly inconsistent; just take $k = 2$ in the second expression and $k = ell = 1$ in the first. There are a few different conventions you could take here, depending on whether you're thinking of exterior powers as a quotient or a subspace of tensor powers. Thinking in terms of quotients is much more convenient (and gets rid of all of these pesky factorial factors) but for some purposes, sometimes people think in terms of subspaces, and then things get messy.
$endgroup$
– Qiaochu Yuan
Jul 31 '16 at 20:52