What is “exterior” about an exterior product?
This is a question about terminology. What is "inner" about an inner product, or "outer" about an outer product?
terminology
asked 2 hours ago
Tobin Fricke
1,457818
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This is a question about terminology. What is "inner" about an inner product, or "outer" about an outer product?
terminology
asked 2 hours ago
Tobin Fricke
1,457818
add a comment |
This is a question about terminology. What is "inner" about an inner product, or "outer" about an outer product?
terminology
asked 2 hours ago
Tobin Fricke
1,457818
This is a question about terminology. What is "inner" about an inner product, or "outer" about an outer product?
terminology
terminology
asked 2 hours ago
Tobin Fricke
1,457818
asked 2 hours ago
Tobin Fricke
1,457818
asked 2 hours ago
Tobin Fricke
1,457818
asked 2 hours ago
Tobin Fricke
1,457818
asked 2 hours ago
Tobin Fricke
1,457818
1,457818
add a comment |
add a comment |
4 Answers
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This terminology (or rather its literal German translation) was introduced by Grassmann.
I named the former product exterior, the latter interior, reflecting that the former was nonzero only when involving independent directions, the latter only when involving a shared, i.e., partly common one.
H. Grassmann, Die lineale Ausdehnungslehre [archive.org] (1844) x-xi.
NB the words in the original text translated in the above excerpt to exterior and interior are respectively äussere and innere, but can also be translated respectively to outer and inner, and now all four terms have distinct meanings in this context.
For more detail, see the answer to essentially the same question on MathOverflow, from where the above translation was lifted (which motivated setting this answer as a community wiki answer):
Etymology of "exterior" in "exterior calculus".
add a comment |
Consider the case n=3. The interior product of two vectors is non-zero when one lives in the span of the other. The outer product of two vectors is non-zero when one lives outside the span of the other.
This terminology can be traced back to one of the earlier German texts on linear algebra but I can't quite recall the name. I'll come back if I recall it.
EDIT: Found my source. Take a look at section 1.2 of this.
answered 2 hours ago
Wraith1995
598315
add a comment |
The exterior (= outer) product takes values in a "higher" dimensional space: if $V$ has dimension $n$ and $v, w in V$ then $v wedge w in Lambda^2(V)$ and that space has dimension $binom{n}{2}$, which is bigger than $n$ when $n > 3$. In Euclidean space, the wedge product of two vectors is represented by a parallelogram with the original vectors as its edges.
The inner/outer terminology goes back to Grassmann.
answered 2 hours ago
KCd
16.5k4075
add a comment |
Well if I take the inner product of two vectors I go from the full vector space down to the base field over which it's defined so its internal in that regard. As for the outer product I think of this in terms of matrix multiplication, if we reverse the inner product $x^Ty$ to $xy^T$ for column vectors in a space of dimension $n$ instead we end up with a matrix which is a larger vector space than the original space with $n^2$ dimensions. Metaphorically I think of this as identifying a vector space with a subspace of a larger space, and the rest of that space is "outer space" in the same way the Earth is in a much larger universe.
answered 2 hours ago
CyclotomicField
2,1821313
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
This terminology (or rather its literal German translation) was introduced by Grassmann.
I named the former product exterior, the latter interior, reflecting that the former was nonzero only when involving independent directions, the latter only when involving a shared, i.e., partly common one.
H. Grassmann, Die lineale Ausdehnungslehre [archive.org] (1844) x-xi.
NB the words in the original text translated in the above excerpt to exterior and interior are respectively äussere and innere, but can also be translated respectively to outer and inner, and now all four terms have distinct meanings in this context.
For more detail, see the answer to essentially the same question on MathOverflow, from where the above translation was lifted (which motivated setting this answer as a community wiki answer):
Etymology of "exterior" in "exterior calculus".
add a comment |
This terminology (or rather its literal German translation) was introduced by Grassmann.
I named the former product exterior, the latter interior, reflecting that the former was nonzero only when involving independent directions, the latter only when involving a shared, i.e., partly common one.
H. Grassmann, Die lineale Ausdehnungslehre [archive.org] (1844) x-xi.
NB the words in the original text translated in the above excerpt to exterior and interior are respectively äussere and innere, but can also be translated respectively to outer and inner, and now all four terms have distinct meanings in this context.
For more detail, see the answer to essentially the same question on MathOverflow, from where the above translation was lifted (which motivated setting this answer as a community wiki answer):
Etymology of "exterior" in "exterior calculus".
add a comment |
This terminology (or rather its literal German translation) was introduced by Grassmann.
I named the former product exterior, the latter interior, reflecting that the former was nonzero only when involving independent directions, the latter only when involving a shared, i.e., partly common one.
H. Grassmann, Die lineale Ausdehnungslehre [archive.org] (1844) x-xi.
NB the words in the original text translated in the above excerpt to exterior and interior are respectively äussere and innere, but can also be translated respectively to outer and inner, and now all four terms have distinct meanings in this context.
For more detail, see the answer to essentially the same question on MathOverflow, from where the above translation was lifted (which motivated setting this answer as a community wiki answer):
Etymology of "exterior" in "exterior calculus".
This terminology (or rather its literal German translation) was introduced by Grassmann.
I named the former product exterior, the latter interior, reflecting that the former was nonzero only when involving independent directions, the latter only when involving a shared, i.e., partly common one.
H. Grassmann, Die lineale Ausdehnungslehre [archive.org] (1844) x-xi.
NB the words in the original text translated in the above excerpt to exterior and interior are respectively äussere and innere, but can also be translated respectively to outer and inner, and now all four terms have distinct meanings in this context.
For more detail, see the answer to essentially the same question on MathOverflow, from where the above translation was lifted (which motivated setting this answer as a community wiki answer):
Etymology of "exterior" in "exterior calculus".
edited 1 hour ago
community wiki
2 revs
Travis
add a comment |
add a comment |
Consider the case n=3. The interior product of two vectors is non-zero when one lives in the span of the other. The outer product of two vectors is non-zero when one lives outside the span of the other.
This terminology can be traced back to one of the earlier German texts on linear algebra but I can't quite recall the name. I'll come back if I recall it.
EDIT: Found my source. Take a look at section 1.2 of this.
answered 2 hours ago
Wraith1995
598315
add a comment |
Consider the case n=3. The interior product of two vectors is non-zero when one lives in the span of the other. The outer product of two vectors is non-zero when one lives outside the span of the other.
This terminology can be traced back to one of the earlier German texts on linear algebra but I can't quite recall the name. I'll come back if I recall it.
EDIT: Found my source. Take a look at section 1.2 of this.
answered 2 hours ago
Wraith1995
598315
add a comment |
Consider the case n=3. The interior product of two vectors is non-zero when one lives in the span of the other. The outer product of two vectors is non-zero when one lives outside the span of the other.
This terminology can be traced back to one of the earlier German texts on linear algebra but I can't quite recall the name. I'll come back if I recall it.
EDIT: Found my source. Take a look at section 1.2 of this.
answered 2 hours ago
Wraith1995
598315
Consider the case n=3. The interior product of two vectors is non-zero when one lives in the span of the other. The outer product of two vectors is non-zero when one lives outside the span of the other.
This terminology can be traced back to one of the earlier German texts on linear algebra but I can't quite recall the name. I'll come back if I recall it.
EDIT: Found my source. Take a look at section 1.2 of this.
answered 2 hours ago
Wraith1995
598315
answered 2 hours ago
Wraith1995
598315
answered 2 hours ago
Wraith1995
598315
answered 2 hours ago
Wraith1995
598315
598315
add a comment |
add a comment |
The exterior (= outer) product takes values in a "higher" dimensional space: if $V$ has dimension $n$ and $v, w in V$ then $v wedge w in Lambda^2(V)$ and that space has dimension $binom{n}{2}$, which is bigger than $n$ when $n > 3$. In Euclidean space, the wedge product of two vectors is represented by a parallelogram with the original vectors as its edges.
The inner/outer terminology goes back to Grassmann.
answered 2 hours ago
KCd
16.5k4075
add a comment |
The exterior (= outer) product takes values in a "higher" dimensional space: if $V$ has dimension $n$ and $v, w in V$ then $v wedge w in Lambda^2(V)$ and that space has dimension $binom{n}{2}$, which is bigger than $n$ when $n > 3$. In Euclidean space, the wedge product of two vectors is represented by a parallelogram with the original vectors as its edges.
The inner/outer terminology goes back to Grassmann.
answered 2 hours ago
KCd
16.5k4075
add a comment |
The exterior (= outer) product takes values in a "higher" dimensional space: if $V$ has dimension $n$ and $v, w in V$ then $v wedge w in Lambda^2(V)$ and that space has dimension $binom{n}{2}$, which is bigger than $n$ when $n > 3$. In Euclidean space, the wedge product of two vectors is represented by a parallelogram with the original vectors as its edges.
The inner/outer terminology goes back to Grassmann.
answered 2 hours ago
KCd
16.5k4075
The exterior (= outer) product takes values in a "higher" dimensional space: if $V$ has dimension $n$ and $v, w in V$ then $v wedge w in Lambda^2(V)$ and that space has dimension $binom{n}{2}$, which is bigger than $n$ when $n > 3$. In Euclidean space, the wedge product of two vectors is represented by a parallelogram with the original vectors as its edges.
The inner/outer terminology goes back to Grassmann.
answered 2 hours ago
KCd
16.5k4075
answered 2 hours ago
KCd
16.5k4075
answered 2 hours ago
KCd
16.5k4075
answered 2 hours ago
KCd
16.5k4075
16.5k4075
add a comment |
add a comment |
Well if I take the inner product of two vectors I go from the full vector space down to the base field over which it's defined so its internal in that regard. As for the outer product I think of this in terms of matrix multiplication, if we reverse the inner product $x^Ty$ to $xy^T$ for column vectors in a space of dimension $n$ instead we end up with a matrix which is a larger vector space than the original space with $n^2$ dimensions. Metaphorically I think of this as identifying a vector space with a subspace of a larger space, and the rest of that space is "outer space" in the same way the Earth is in a much larger universe.
answered 2 hours ago
CyclotomicField
2,1821313
add a comment |
Well if I take the inner product of two vectors I go from the full vector space down to the base field over which it's defined so its internal in that regard. As for the outer product I think of this in terms of matrix multiplication, if we reverse the inner product $x^Ty$ to $xy^T$ for column vectors in a space of dimension $n$ instead we end up with a matrix which is a larger vector space than the original space with $n^2$ dimensions. Metaphorically I think of this as identifying a vector space with a subspace of a larger space, and the rest of that space is "outer space" in the same way the Earth is in a much larger universe.
answered 2 hours ago
CyclotomicField
2,1821313
add a comment |
Well if I take the inner product of two vectors I go from the full vector space down to the base field over which it's defined so its internal in that regard. As for the outer product I think of this in terms of matrix multiplication, if we reverse the inner product $x^Ty$ to $xy^T$ for column vectors in a space of dimension $n$ instead we end up with a matrix which is a larger vector space than the original space with $n^2$ dimensions. Metaphorically I think of this as identifying a vector space with a subspace of a larger space, and the rest of that space is "outer space" in the same way the Earth is in a much larger universe.
answered 2 hours ago
CyclotomicField
2,1821313
Well if I take the inner product of two vectors I go from the full vector space down to the base field over which it's defined so its internal in that regard. As for the outer product I think of this in terms of matrix multiplication, if we reverse the inner product $x^Ty$ to $xy^T$ for column vectors in a space of dimension $n$ instead we end up with a matrix which is a larger vector space than the original space with $n^2$ dimensions. Metaphorically I think of this as identifying a vector space with a subspace of a larger space, and the rest of that space is "outer space" in the same way the Earth is in a much larger universe.
answered 2 hours ago
CyclotomicField
2,1821313
answered 2 hours ago
CyclotomicField
2,1821313
answered 2 hours ago
CyclotomicField
2,1821313
answered 2 hours ago
CyclotomicField
2,1821313
2,1821313
add a comment |
add a comment |
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