Mixing
How can we define a covariant $k$-tensor this way if tensors are already defined by the tensor product?
$begingroup$
From Lee's Intro to Smooth Manifolds:
If we take $k$ covectors, $varepsilon^{i_1}, dots, varepsilon^{i_k},$ then the tensor product is defined by $$varepsilon^{i_1}otimes cdots otimes varepsilon^{i_k}(v_1, dots, v_k)=varepsilon^{i_1}(v_1)cdots varepsilon^{i_k}(v_k).$$
So, how can we define a covariant $k$-tensor as in the image above?
linear-algebra abstract-algebra definition tensor-products
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
$endgroup$
add a comment |
$begingroup$
From Lee's Intro to Smooth Manifolds:
If we take $k$ covectors, $varepsilon^{i_1}, dots, varepsilon^{i_k},$ then the tensor product is defined by $$varepsilon^{i_1}otimes cdots otimes varepsilon^{i_k}(v_1, dots, v_k)=varepsilon^{i_1}(v_1)cdots varepsilon^{i_k}(v_k).$$
So, how can we define a covariant $k$-tensor as in the image above?
linear-algebra abstract-algebra definition tensor-products
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
$endgroup$
$begingroup$
As you've noticed, $epsilon^I$ is not the tensor product of $epsilon^{i_1}cdots epsilon^{i_k}$, however it is a multilinear map from $V$ to $k$, and thus corresponds to an element of $V^{*otimes k}$.
$endgroup$
– jgon
Jan 14 at 23:53
$begingroup$
Why was this question downvoted? If the question is obvious to you, it may not be obvious to others.
$endgroup$
– Al Jebr
Jan 17 at 21:14
2
$begingroup$
while I am not the downvoter, I figured I'd offer my thoughts, since my earlier comment resulted in me receiving a notification about your comment. I suspect the reason this question was downvoted is that it doesn't show very much research or effort. After all, you posted an image with two sentences that don't really explain what you found confusing. You also ignored my comment (from 3 days ago) attempting to answer your question (which may have factored into a negative perception of your question, though I'm unbothered by it).
$endgroup$
– jgon
Jan 17 at 21:24
add a comment |
$begingroup$
From Lee's Intro to Smooth Manifolds:
If we take $k$ covectors, $varepsilon^{i_1}, dots, varepsilon^{i_k},$ then the tensor product is defined by $$varepsilon^{i_1}otimes cdots otimes varepsilon^{i_k}(v_1, dots, v_k)=varepsilon^{i_1}(v_1)cdots varepsilon^{i_k}(v_k).$$
So, how can we define a covariant $k$-tensor as in the image above?
linear-algebra abstract-algebra definition tensor-products
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
$endgroup$
From Lee's Intro to Smooth Manifolds:
If we take $k$ covectors, $varepsilon^{i_1}, dots, varepsilon^{i_k},$ then the tensor product is defined by $$varepsilon^{i_1}otimes cdots otimes varepsilon^{i_k}(v_1, dots, v_k)=varepsilon^{i_1}(v_1)cdots varepsilon^{i_k}(v_k).$$
So, how can we define a covariant $k$-tensor as in the image above?
linear-algebra abstract-algebra definition tensor-products
linear-algebra abstract-algebra definition tensor-products
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
asked Jan 14 at 16:44
Al JebrAl Jebr
4,25343277
4,25343277
$begingroup$
As you've noticed, $epsilon^I$ is not the tensor product of $epsilon^{i_1}cdots epsilon^{i_k}$, however it is a multilinear map from $V$ to $k$, and thus corresponds to an element of $V^{*otimes k}$.
$endgroup$
– jgon
Jan 14 at 23:53
$begingroup$
Why was this question downvoted? If the question is obvious to you, it may not be obvious to others.
$endgroup$
– Al Jebr
Jan 17 at 21:14
2
$begingroup$
while I am not the downvoter, I figured I'd offer my thoughts, since my earlier comment resulted in me receiving a notification about your comment. I suspect the reason this question was downvoted is that it doesn't show very much research or effort. After all, you posted an image with two sentences that don't really explain what you found confusing. You also ignored my comment (from 3 days ago) attempting to answer your question (which may have factored into a negative perception of your question, though I'm unbothered by it).
$endgroup$
– jgon
Jan 17 at 21:24
add a comment |
$begingroup$
As you've noticed, $epsilon^I$ is not the tensor product of $epsilon^{i_1}cdots epsilon^{i_k}$, however it is a multilinear map from $V$ to $k$, and thus corresponds to an element of $V^{*otimes k}$.
$endgroup$
– jgon
Jan 14 at 23:53
$begingroup$
Why was this question downvoted? If the question is obvious to you, it may not be obvious to others.
$endgroup$
– Al Jebr
Jan 17 at 21:14
2
$begingroup$
while I am not the downvoter, I figured I'd offer my thoughts, since my earlier comment resulted in me receiving a notification about your comment. I suspect the reason this question was downvoted is that it doesn't show very much research or effort. After all, you posted an image with two sentences that don't really explain what you found confusing. You also ignored my comment (from 3 days ago) attempting to answer your question (which may have factored into a negative perception of your question, though I'm unbothered by it).
$endgroup$
– jgon
Jan 17 at 21:24
$begingroup$
As you've noticed, $epsilon^I$ is not the tensor product of $epsilon^{i_1}cdots epsilon^{i_k}$, however it is a multilinear map from $V$ to $k$, and thus corresponds to an element of $V^{*otimes k}$.
$endgroup$
– jgon
Jan 14 at 23:53
$begingroup$
As you've noticed, $epsilon^I$ is not the tensor product of $epsilon^{i_1}cdots epsilon^{i_k}$, however it is a multilinear map from $V$ to $k$, and thus corresponds to an element of $V^{*otimes k}$.
$endgroup$
– jgon
Jan 14 at 23:53
$begingroup$
Why was this question downvoted? If the question is obvious to you, it may not be obvious to others.
$endgroup$
– Al Jebr
Jan 17 at 21:14
$begingroup$
Why was this question downvoted? If the question is obvious to you, it may not be obvious to others.
$endgroup$
– Al Jebr
Jan 17 at 21:14
2
2
$begingroup$
while I am not the downvoter, I figured I'd offer my thoughts, since my earlier comment resulted in me receiving a notification about your comment. I suspect the reason this question was downvoted is that it doesn't show very much research or effort. After all, you posted an image with two sentences that don't really explain what you found confusing. You also ignored my comment (from 3 days ago) attempting to answer your question (which may have factored into a negative perception of your question, though I'm unbothered by it).
$endgroup$
– jgon
Jan 17 at 21:24
$begingroup$
while I am not the downvoter, I figured I'd offer my thoughts, since my earlier comment resulted in me receiving a notification about your comment. I suspect the reason this question was downvoted is that it doesn't show very much research or effort. After all, you posted an image with two sentences that don't really explain what you found confusing. You also ignored my comment (from 3 days ago) attempting to answer your question (which may have factored into a negative perception of your question, though I'm unbothered by it).
$endgroup$
– jgon
Jan 17 at 21:24
add a comment |
1 Answer
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oldest
votes
$begingroup$
You define it as shown, as there is no conflict of definitions. Perhaps the $k = 2$ case can render a bit more readability.
Although
$$ varepsilon ^ {i_1} otimes varepsilon ^ {i_2}, varepsilon^{(i_1,i_2)}:V^2rightarrow mathbb{R}$$
have the same domain and codomain, they are defined entirely differently, as shown in your post:
$$varepsilon^{i_1}otimesvarepsilon^{i_2}(v_1,v_2) = varepsilon^{i_1}(v_1) varepsilon^{i_2}(v_2)$$
$$varepsilon^{(i_1,i_2)}(v_1,v_2) = varepsilon^{i_1}(v_1)varepsilon^{i_2}(v_2) - varepsilon^{i_1}(v_2)varepsilon^{i_2}(v_1)$$
answered Jan 15 at 1:36
MetricMetric
1,23649
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
You define it as shown, as there is no conflict of definitions. Perhaps the $k = 2$ case can render a bit more readability.
Although
$$ varepsilon ^ {i_1} otimes varepsilon ^ {i_2}, varepsilon^{(i_1,i_2)}:V^2rightarrow mathbb{R}$$
have the same domain and codomain, they are defined entirely differently, as shown in your post:
$$varepsilon^{i_1}otimesvarepsilon^{i_2}(v_1,v_2) = varepsilon^{i_1}(v_1) varepsilon^{i_2}(v_2)$$
$$varepsilon^{(i_1,i_2)}(v_1,v_2) = varepsilon^{i_1}(v_1)varepsilon^{i_2}(v_2) - varepsilon^{i_1}(v_2)varepsilon^{i_2}(v_1)$$
answered Jan 15 at 1:36
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
You define it as shown, as there is no conflict of definitions. Perhaps the $k = 2$ case can render a bit more readability.
Although
$$ varepsilon ^ {i_1} otimes varepsilon ^ {i_2}, varepsilon^{(i_1,i_2)}:V^2rightarrow mathbb{R}$$
have the same domain and codomain, they are defined entirely differently, as shown in your post:
$$varepsilon^{i_1}otimesvarepsilon^{i_2}(v_1,v_2) = varepsilon^{i_1}(v_1) varepsilon^{i_2}(v_2)$$
$$varepsilon^{(i_1,i_2)}(v_1,v_2) = varepsilon^{i_1}(v_1)varepsilon^{i_2}(v_2) - varepsilon^{i_1}(v_2)varepsilon^{i_2}(v_1)$$
answered Jan 15 at 1:36
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
You define it as shown, as there is no conflict of definitions. Perhaps the $k = 2$ case can render a bit more readability.
Although
$$ varepsilon ^ {i_1} otimes varepsilon ^ {i_2}, varepsilon^{(i_1,i_2)}:V^2rightarrow mathbb{R}$$
have the same domain and codomain, they are defined entirely differently, as shown in your post:
$$varepsilon^{i_1}otimesvarepsilon^{i_2}(v_1,v_2) = varepsilon^{i_1}(v_1) varepsilon^{i_2}(v_2)$$
$$varepsilon^{(i_1,i_2)}(v_1,v_2) = varepsilon^{i_1}(v_1)varepsilon^{i_2}(v_2) - varepsilon^{i_1}(v_2)varepsilon^{i_2}(v_1)$$
answered Jan 15 at 1:36
MetricMetric
1,23649
$endgroup$
You define it as shown, as there is no conflict of definitions. Perhaps the $k = 2$ case can render a bit more readability.
Although
$$ varepsilon ^ {i_1} otimes varepsilon ^ {i_2}, varepsilon^{(i_1,i_2)}:V^2rightarrow mathbb{R}$$
have the same domain and codomain, they are defined entirely differently, as shown in your post:
$$varepsilon^{i_1}otimesvarepsilon^{i_2}(v_1,v_2) = varepsilon^{i_1}(v_1) varepsilon^{i_2}(v_2)$$
$$varepsilon^{(i_1,i_2)}(v_1,v_2) = varepsilon^{i_1}(v_1)varepsilon^{i_2}(v_2) - varepsilon^{i_1}(v_2)varepsilon^{i_2}(v_1)$$
answered Jan 15 at 1:36
MetricMetric
1,23649
answered Jan 15 at 1:36
MetricMetric
1,23649
answered Jan 15 at 1:36
MetricMetric
1,23649
answered Jan 15 at 1:36
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
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$begingroup$
As you've noticed, $epsilon^I$ is not the tensor product of $epsilon^{i_1}cdots epsilon^{i_k}$, however it is a multilinear map from $V$ to $k$, and thus corresponds to an element of $V^{*otimes k}$.
$endgroup$
– jgon
Jan 14 at 23:53
$begingroup$
Why was this question downvoted? If the question is obvious to you, it may not be obvious to others.
$endgroup$
– Al Jebr
Jan 17 at 21:14
2
$begingroup$
while I am not the downvoter, I figured I'd offer my thoughts, since my earlier comment resulted in me receiving a notification about your comment. I suspect the reason this question was downvoted is that it doesn't show very much research or effort. After all, you posted an image with two sentences that don't really explain what you found confusing. You also ignored my comment (from 3 days ago) attempting to answer your question (which may have factored into a negative perception of your question, though I'm unbothered by it).
$endgroup$
– jgon
Jan 17 at 21:24