Are all bivectors in three dimensions simple?
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I want to show that all bivectors in three dimensions are simple.
If I understand correctly, a bivector is simply an element from the two-fold exterior product $bigwedge^2V$ of a vector space $V$, right?
We can define $wedge(e_iotimes e_j):=e_iotimes e_j - e_jotimes e_i$. Now let $T=t^{ij}e_iwedge e_jmapsto t^{ij}(e_iotimes e_j - e_jotimes e_i)=(t^{ij}-t^{ji})e_iotimes e_j$. This is injective. Because the wedge product as a linear map from the tensor product to the exterior product maps all symmetric tensors to 0.
We see that total antisymmetric tensors in this case are represented by skew-symmetric matrices. To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
But the rank of a skew-symmetric matrix is never one.
I must have made a conceptual mistake somewhere again. Does anybody have a hint for me?
Geometrically, one can use the canonical isomorphism between the two-fold exterior product and $mathbb{R}^3$ itself to show that any anyisymmetric tensor can be thought of as a vector in 3D, which in turn can be represented by the cross product of two vectors, that are not collinear to each other but orthogonal to that vector.
vector-spaces tensor-products tensors exterior-algebra
asked Nov 13 at 22:59
Thomas Wening
12110
add a comment |
up vote
1
down vote
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I want to show that all bivectors in three dimensions are simple.
If I understand correctly, a bivector is simply an element from the two-fold exterior product $bigwedge^2V$ of a vector space $V$, right?
We can define $wedge(e_iotimes e_j):=e_iotimes e_j - e_jotimes e_i$. Now let $T=t^{ij}e_iwedge e_jmapsto t^{ij}(e_iotimes e_j - e_jotimes e_i)=(t^{ij}-t^{ji})e_iotimes e_j$. This is injective. Because the wedge product as a linear map from the tensor product to the exterior product maps all symmetric tensors to 0.
We see that total antisymmetric tensors in this case are represented by skew-symmetric matrices. To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
But the rank of a skew-symmetric matrix is never one.
I must have made a conceptual mistake somewhere again. Does anybody have a hint for me?
Geometrically, one can use the canonical isomorphism between the two-fold exterior product and $mathbb{R}^3$ itself to show that any anyisymmetric tensor can be thought of as a vector in 3D, which in turn can be represented by the cross product of two vectors, that are not collinear to each other but orthogonal to that vector.
vector-spaces tensor-products tensors exterior-algebra
asked Nov 13 at 22:59
Thomas Wening
12110
2
$(e_1+e_2)wedge(e_3-e_1)$
– user8268
Nov 13 at 23:12
1
Does "simple" mean "of the form $u wedge v$ for some $u, v in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n geq 2$, then each element of $Lambda^{n-1}left(Vright)$ can be written as $v_1 wedge v_2 wedge cdots wedge v_{n-1}$ for some $v_1, v_2, ldots, v_{n-1} in V$.
– darij grinberg
Nov 14 at 2:42
Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $mathbb{R} ^3$ itself.
– Thomas Wening
Nov 14 at 11:05
Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post.
– Thomas Wening
Nov 14 at 14:24
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show that all bivectors in three dimensions are simple.
If I understand correctly, a bivector is simply an element from the two-fold exterior product $bigwedge^2V$ of a vector space $V$, right?
We can define $wedge(e_iotimes e_j):=e_iotimes e_j - e_jotimes e_i$. Now let $T=t^{ij}e_iwedge e_jmapsto t^{ij}(e_iotimes e_j - e_jotimes e_i)=(t^{ij}-t^{ji})e_iotimes e_j$. This is injective. Because the wedge product as a linear map from the tensor product to the exterior product maps all symmetric tensors to 0.
We see that total antisymmetric tensors in this case are represented by skew-symmetric matrices. To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
But the rank of a skew-symmetric matrix is never one.
I must have made a conceptual mistake somewhere again. Does anybody have a hint for me?
Geometrically, one can use the canonical isomorphism between the two-fold exterior product and $mathbb{R}^3$ itself to show that any anyisymmetric tensor can be thought of as a vector in 3D, which in turn can be represented by the cross product of two vectors, that are not collinear to each other but orthogonal to that vector.
vector-spaces tensor-products tensors exterior-algebra
asked Nov 13 at 22:59
Thomas Wening
12110
I want to show that all bivectors in three dimensions are simple.
If I understand correctly, a bivector is simply an element from the two-fold exterior product $bigwedge^2V$ of a vector space $V$, right?
We can define $wedge(e_iotimes e_j):=e_iotimes e_j - e_jotimes e_i$. Now let $T=t^{ij}e_iwedge e_jmapsto t^{ij}(e_iotimes e_j - e_jotimes e_i)=(t^{ij}-t^{ji})e_iotimes e_j$. This is injective. Because the wedge product as a linear map from the tensor product to the exterior product maps all symmetric tensors to 0.
We see that total antisymmetric tensors in this case are represented by skew-symmetric matrices. To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
But the rank of a skew-symmetric matrix is never one.
I must have made a conceptual mistake somewhere again. Does anybody have a hint for me?
Geometrically, one can use the canonical isomorphism between the two-fold exterior product and $mathbb{R}^3$ itself to show that any anyisymmetric tensor can be thought of as a vector in 3D, which in turn can be represented by the cross product of two vectors, that are not collinear to each other but orthogonal to that vector.
vector-spaces tensor-products tensors exterior-algebra
vector-spaces tensor-products tensors exterior-algebra
asked Nov 13 at 22:59
Thomas Wening
12110
asked Nov 13 at 22:59
Thomas Wening
12110
edited Nov 14 at 14:37
asked Nov 13 at 22:59
Thomas Wening
12110
asked Nov 13 at 22:59
Thomas Wening
12110
asked Nov 13 at 22:59
Thomas Wening
12110
12110
2
$(e_1+e_2)wedge(e_3-e_1)$
– user8268
Nov 13 at 23:12
1
Does "simple" mean "of the form $u wedge v$ for some $u, v in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n geq 2$, then each element of $Lambda^{n-1}left(Vright)$ can be written as $v_1 wedge v_2 wedge cdots wedge v_{n-1}$ for some $v_1, v_2, ldots, v_{n-1} in V$.
– darij grinberg
Nov 14 at 2:42
Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $mathbb{R} ^3$ itself.
– Thomas Wening
Nov 14 at 11:05
Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post.
– Thomas Wening
Nov 14 at 14:24
add a comment |
2
$(e_1+e_2)wedge(e_3-e_1)$
– user8268
Nov 13 at 23:12
1
Does "simple" mean "of the form $u wedge v$ for some $u, v in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n geq 2$, then each element of $Lambda^{n-1}left(Vright)$ can be written as $v_1 wedge v_2 wedge cdots wedge v_{n-1}$ for some $v_1, v_2, ldots, v_{n-1} in V$.
– darij grinberg
Nov 14 at 2:42
Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $mathbb{R} ^3$ itself.
– Thomas Wening
Nov 14 at 11:05
Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post.
– Thomas Wening
Nov 14 at 14:24
2
2
$(e_1+e_2)wedge(e_3-e_1)$
– user8268
Nov 13 at 23:12
$(e_1+e_2)wedge(e_3-e_1)$
– user8268
Nov 13 at 23:12
1
1
Does "simple" mean "of the form $u wedge v$ for some $u, v in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n geq 2$, then each element of $Lambda^{n-1}left(Vright)$ can be written as $v_1 wedge v_2 wedge cdots wedge v_{n-1}$ for some $v_1, v_2, ldots, v_{n-1} in V$.
– darij grinberg
Nov 14 at 2:42
Does "simple" mean "of the form $u wedge v$ for some $u, v in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n geq 2$, then each element of $Lambda^{n-1}left(Vright)$ can be written as $v_1 wedge v_2 wedge cdots wedge v_{n-1}$ for some $v_1, v_2, ldots, v_{n-1} in V$.
– darij grinberg
Nov 14 at 2:42
Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $mathbb{R} ^3$ itself.
– Thomas Wening
Nov 14 at 11:05
Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $mathbb{R} ^3$ itself.
– Thomas Wening
Nov 14 at 11:05
Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post.
– Thomas Wening
Nov 14 at 14:24
Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post.
– Thomas Wening
Nov 14 at 14:24
add a comment |
1 Answer
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1
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To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
This doesn't follow, since as you say the rank of a skew-symmetric matrix can never be $1$. You're conflating $e_i otimes e_j - e_j otimes e_i$, which as a tensor has rank $2$ or $0$, with $e_i wedge e_j$. These are not the same object; one of them lives in $V^{otimes 2}$ and the other one lives in $Lambda^2(V)$. In general I don't recommend thinking in terms of antisymmetric tensors; it makes the exterior product look much more complicated than it is.
Anyway, here's a proof of darij's more general claim in the comments. Let $v_1 in Lambda^{n-1}(V)$ be a vector, where $dim V = n$. Choose a nonzero element $omega in Lambda^n(V)$, hence an identification of it with the ground field $k$. Then the exterior product
$$wedge : V times Lambda^{n-1}(V) to Lambda^n(V) cong k$$
is a nondegenerate bilinear pairing. Extend $v_1$ to a basis $v_1, dots v_n in Lambda^{n-1}(V)$. Then it has a unique dual basis $e_1, dots e_n in V$ defined by the condition that
$$e_i wedge v_j = delta_{ij} omega in Lambda^n(V).$$
Then the $v_i$ must also be the dual basis of the $e_i$ with respect to this pairing. But this dual basis in turn must be
$$v_i = (-1)^{i-1} frac{omega}{e_1 wedge dots wedge e_n} e_1 wedge dots wedge widehat{e_i} wedge dots wedge e_n$$
where the hat denotes that we omit $e_i$, and in particular
$$v_1 = frac{omega}{e_1 wedge dots wedge e_n} e_2 wedge dots wedge e_n.$$
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
1
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
|
show 4 more comments
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
This doesn't follow, since as you say the rank of a skew-symmetric matrix can never be $1$. You're conflating $e_i otimes e_j - e_j otimes e_i$, which as a tensor has rank $2$ or $0$, with $e_i wedge e_j$. These are not the same object; one of them lives in $V^{otimes 2}$ and the other one lives in $Lambda^2(V)$. In general I don't recommend thinking in terms of antisymmetric tensors; it makes the exterior product look much more complicated than it is.
Anyway, here's a proof of darij's more general claim in the comments. Let $v_1 in Lambda^{n-1}(V)$ be a vector, where $dim V = n$. Choose a nonzero element $omega in Lambda^n(V)$, hence an identification of it with the ground field $k$. Then the exterior product
$$wedge : V times Lambda^{n-1}(V) to Lambda^n(V) cong k$$
is a nondegenerate bilinear pairing. Extend $v_1$ to a basis $v_1, dots v_n in Lambda^{n-1}(V)$. Then it has a unique dual basis $e_1, dots e_n in V$ defined by the condition that
$$e_i wedge v_j = delta_{ij} omega in Lambda^n(V).$$
Then the $v_i$ must also be the dual basis of the $e_i$ with respect to this pairing. But this dual basis in turn must be
$$v_i = (-1)^{i-1} frac{omega}{e_1 wedge dots wedge e_n} e_1 wedge dots wedge widehat{e_i} wedge dots wedge e_n$$
where the hat denotes that we omit $e_i$, and in particular
$$v_1 = frac{omega}{e_1 wedge dots wedge e_n} e_2 wedge dots wedge e_n.$$
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
1
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
|
show 4 more comments
up vote
1
down vote
accepted
To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
This doesn't follow, since as you say the rank of a skew-symmetric matrix can never be $1$. You're conflating $e_i otimes e_j - e_j otimes e_i$, which as a tensor has rank $2$ or $0$, with $e_i wedge e_j$. These are not the same object; one of them lives in $V^{otimes 2}$ and the other one lives in $Lambda^2(V)$. In general I don't recommend thinking in terms of antisymmetric tensors; it makes the exterior product look much more complicated than it is.
Anyway, here's a proof of darij's more general claim in the comments. Let $v_1 in Lambda^{n-1}(V)$ be a vector, where $dim V = n$. Choose a nonzero element $omega in Lambda^n(V)$, hence an identification of it with the ground field $k$. Then the exterior product
$$wedge : V times Lambda^{n-1}(V) to Lambda^n(V) cong k$$
is a nondegenerate bilinear pairing. Extend $v_1$ to a basis $v_1, dots v_n in Lambda^{n-1}(V)$. Then it has a unique dual basis $e_1, dots e_n in V$ defined by the condition that
$$e_i wedge v_j = delta_{ij} omega in Lambda^n(V).$$
Then the $v_i$ must also be the dual basis of the $e_i$ with respect to this pairing. But this dual basis in turn must be
$$v_i = (-1)^{i-1} frac{omega}{e_1 wedge dots wedge e_n} e_1 wedge dots wedge widehat{e_i} wedge dots wedge e_n$$
where the hat denotes that we omit $e_i$, and in particular
$$v_1 = frac{omega}{e_1 wedge dots wedge e_n} e_2 wedge dots wedge e_n.$$
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
1
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
|
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
This doesn't follow, since as you say the rank of a skew-symmetric matrix can never be $1$. You're conflating $e_i otimes e_j - e_j otimes e_i$, which as a tensor has rank $2$ or $0$, with $e_i wedge e_j$. These are not the same object; one of them lives in $V^{otimes 2}$ and the other one lives in $Lambda^2(V)$. In general I don't recommend thinking in terms of antisymmetric tensors; it makes the exterior product look much more complicated than it is.
Anyway, here's a proof of darij's more general claim in the comments. Let $v_1 in Lambda^{n-1}(V)$ be a vector, where $dim V = n$. Choose a nonzero element $omega in Lambda^n(V)$, hence an identification of it with the ground field $k$. Then the exterior product
$$wedge : V times Lambda^{n-1}(V) to Lambda^n(V) cong k$$
is a nondegenerate bilinear pairing. Extend $v_1$ to a basis $v_1, dots v_n in Lambda^{n-1}(V)$. Then it has a unique dual basis $e_1, dots e_n in V$ defined by the condition that
$$e_i wedge v_j = delta_{ij} omega in Lambda^n(V).$$
Then the $v_i$ must also be the dual basis of the $e_i$ with respect to this pairing. But this dual basis in turn must be
$$v_i = (-1)^{i-1} frac{omega}{e_1 wedge dots wedge e_n} e_1 wedge dots wedge widehat{e_i} wedge dots wedge e_n$$
where the hat denotes that we omit $e_i$, and in particular
$$v_1 = frac{omega}{e_1 wedge dots wedge e_n} e_2 wedge dots wedge e_n.$$
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
To show that they are all simple, I would have to show that the rank of any 3x3 skew-symmetric matrix is 1.
This doesn't follow, since as you say the rank of a skew-symmetric matrix can never be $1$. You're conflating $e_i otimes e_j - e_j otimes e_i$, which as a tensor has rank $2$ or $0$, with $e_i wedge e_j$. These are not the same object; one of them lives in $V^{otimes 2}$ and the other one lives in $Lambda^2(V)$. In general I don't recommend thinking in terms of antisymmetric tensors; it makes the exterior product look much more complicated than it is.
Anyway, here's a proof of darij's more general claim in the comments. Let $v_1 in Lambda^{n-1}(V)$ be a vector, where $dim V = n$. Choose a nonzero element $omega in Lambda^n(V)$, hence an identification of it with the ground field $k$. Then the exterior product
$$wedge : V times Lambda^{n-1}(V) to Lambda^n(V) cong k$$
is a nondegenerate bilinear pairing. Extend $v_1$ to a basis $v_1, dots v_n in Lambda^{n-1}(V)$. Then it has a unique dual basis $e_1, dots e_n in V$ defined by the condition that
$$e_i wedge v_j = delta_{ij} omega in Lambda^n(V).$$
Then the $v_i$ must also be the dual basis of the $e_i$ with respect to this pairing. But this dual basis in turn must be
$$v_i = (-1)^{i-1} frac{omega}{e_1 wedge dots wedge e_n} e_1 wedge dots wedge widehat{e_i} wedge dots wedge e_n$$
where the hat denotes that we omit $e_i$, and in particular
$$v_1 = frac{omega}{e_1 wedge dots wedge e_n} e_2 wedge dots wedge e_n.$$
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
answered Nov 14 at 21:18
Qiaochu Yuan
274k32578914
274k32578914
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
1
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
|
show 4 more comments
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
1
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
Okay, this calms my mind. I was already worried I had gotten something really wrong about the theorem about the rank and simpleness. Now to your proof: I have a question. What does $frac{omega}{e^1wedgecdotswedge e_n}$ mean?
– Thomas Wening
Nov 14 at 21:41
1
1
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
@Thomas: $Lambda^n(V)$ is one-dimensional, and $omega$ and $e_1 wedge dots wedge e_n$ are two nonzero elements of it, so there's a unique nonzero scalar $c$ such that $omega = c e_1 wedge dots wedge e_n$. It refers to this scalar.
– Qiaochu Yuan
Nov 14 at 21:44
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Okay, I do understand the dual base part. The factor $(-1)^{i-1}$ is needed because you have to permute $e_i$ $i-1$ times in the wedge product $e_iwedge v_i$ to get all the indices in monotonously increasing order, right?
– Thomas Wening
Nov 14 at 21:59
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Yes, that's right.
– Qiaochu Yuan
Nov 14 at 22:00
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
Good. The pairing allows for the definition of a dual basis, once I have fixed one in $bigwedge^2mathbb{R}^3$. But how does that allow me to decompose an arbitrary $omegainbigwedge^{n-1}V$ into $omega=vwedge u$ with $u,vin V$?
– Thomas Wening
Nov 14 at 22:11
|
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2
$(e_1+e_2)wedge(e_3-e_1)$
– user8268
Nov 13 at 23:12
1
Does "simple" mean "of the form $u wedge v$ for some $u, v in V$"? In this case, this is not hard to prove. More generally, if $V$ is an $n$-dimensional vector space over a field with $n geq 2$, then each element of $Lambda^{n-1}left(Vright)$ can be written as $v_1 wedge v_2 wedge cdots wedge v_{n-1}$ for some $v_1, v_2, ldots, v_{n-1} in V$.
– darij grinberg
Nov 14 at 2:42
Okay, I see how the vector given by user8268 decomposes into the one I wrote down. Now the question is how do I arrive at the result given by darij grinberg. I have no clue. I have an interesting idea, though. For 3 dimensions the 2-fold exterior algebra is obviously isomorphic to $mathbb{R} ^3$ itself.
– Thomas Wening
Nov 14 at 11:05
Okay, it turns out that the rank of the coefficient matrix must be equal to 1. I have just gone through the proof. But now I have another problem. It's amended in the post.
– Thomas Wening
Nov 14 at 14:24