Mixing
(Exterior Algebra) Relation between positive oriented space and $r$-forms space.
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Let $S$ a oriented vector space of dimension $m$ and equipped with a inner product. Given $v in S$, let $omega = phi(v) in mathcalA_m-1(S)$ defined by
$$omega(v_1,...,v_m-1) = langle v, v_1,...,v_m-1 rangle.$$
Show that the aplication $phi: S to mathcalA_m-1(S)$ is an isomorphism. If $v in S$ is the first element of a orthonormal positive basis such that the dual is $ e_1,...,e_m $ then $phi(v) = e_2 wedge dots wedge e_m$. Conclude that if $dim S = m$ then every $omega in mathcalA_r(S)$ is decomposable.
Notation. $mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
Idea. First, $dim mathcalA_m-1(S) = binommm-1 = m$ and, since $phi$ is linear, we need to show that $ker phi = 0 $. Note that
$$phi(v) = 0 Longleftrightarrow omega = 0 Longleftrightarrow langle v, v_1,...v_m-1 rangle = 0 Longleftrightarrow v = 0.$$
Is this right?
I couldn't prove the second part. I would like help. I didn't try to make the conclusion yet, so I wanted help only in the second part.
linear-algebra multilinear-algebra exterior-algebra
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
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up vote
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Let $S$ a oriented vector space of dimension $m$ and equipped with a inner product. Given $v in S$, let $omega = phi(v) in mathcalA_m-1(S)$ defined by
$$omega(v_1,...,v_m-1) = langle v, v_1,...,v_m-1 rangle.$$
Show that the aplication $phi: S to mathcalA_m-1(S)$ is an isomorphism. If $v in S$ is the first element of a orthonormal positive basis such that the dual is $ e_1,...,e_m $ then $phi(v) = e_2 wedge dots wedge e_m$. Conclude that if $dim S = m$ then every $omega in mathcalA_r(S)$ is decomposable.
Notation. $mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
Idea. First, $dim mathcalA_m-1(S) = binommm-1 = m$ and, since $phi$ is linear, we need to show that $ker phi = 0 $. Note that
$$phi(v) = 0 Longleftrightarrow omega = 0 Longleftrightarrow langle v, v_1,...v_m-1 rangle = 0 Longleftrightarrow v = 0.$$
Is this right?
I couldn't prove the second part. I would like help. I didn't try to make the conclusion yet, so I wanted help only in the second part.
linear-algebra multilinear-algebra exterior-algebra
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $S$ a oriented vector space of dimension $m$ and equipped with a inner product. Given $v in S$, let $omega = phi(v) in mathcalA_m-1(S)$ defined by
$$omega(v_1,...,v_m-1) = langle v, v_1,...,v_m-1 rangle.$$
Show that the aplication $phi: S to mathcalA_m-1(S)$ is an isomorphism. If $v in S$ is the first element of a orthonormal positive basis such that the dual is $ e_1,...,e_m $ then $phi(v) = e_2 wedge dots wedge e_m$. Conclude that if $dim S = m$ then every $omega in mathcalA_r(S)$ is decomposable.
Notation. $mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
Idea. First, $dim mathcalA_m-1(S) = binommm-1 = m$ and, since $phi$ is linear, we need to show that $ker phi = 0 $. Note that
$$phi(v) = 0 Longleftrightarrow omega = 0 Longleftrightarrow langle v, v_1,...v_m-1 rangle = 0 Longleftrightarrow v = 0.$$
Is this right?
I couldn't prove the second part. I would like help. I didn't try to make the conclusion yet, so I wanted help only in the second part.
linear-algebra multilinear-algebra exterior-algebra
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
Let $S$ a oriented vector space of dimension $m$ and equipped with a inner product. Given $v in S$, let $omega = phi(v) in mathcalA_m-1(S)$ defined by
$$omega(v_1,...,v_m-1) = langle v, v_1,...,v_m-1 rangle.$$
Show that the aplication $phi: S to mathcalA_m-1(S)$ is an isomorphism. If $v in S$ is the first element of a orthonormal positive basis such that the dual is $ e_1,...,e_m $ then $phi(v) = e_2 wedge dots wedge e_m$. Conclude that if $dim S = m$ then every $omega in mathcalA_r(S)$ is decomposable.
Notation. $mathcalA_r(S)$ denotes the vector space of $r$-linear forms over $S$.
Idea. First, $dim mathcalA_m-1(S) = binommm-1 = m$ and, since $phi$ is linear, we need to show that $ker phi = 0 $. Note that
$$phi(v) = 0 Longleftrightarrow omega = 0 Longleftrightarrow langle v, v_1,...v_m-1 rangle = 0 Longleftrightarrow v = 0.$$
Is this right?
I couldn't prove the second part. I would like help. I didn't try to make the conclusion yet, so I wanted help only in the second part.
linear-algebra multilinear-algebra exterior-algebra
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
asked Jul 17 at 5:35
Lucas Corrêa
1,202319
1,202319
add a comment |Â
add a comment |Â
1 Answer
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Your idea is indeed a correct proof; maybe in the third equivalent statement in the highlighted row one could insert "for all $v_1, ldots, v_m-1$".
For the second statement, it suffices to show that the two $m-1$-forms coincide on the given orthonormal positive basis, which we denote by $v = b_1, b_2, ldots, b_n$; but indeed,
$$
phi(v)(b_1, ldots, hatb_k, ldots, b_n) = langle b_1, b_1, ldots, hatb_k, ldots, b_n rangle = begincases
+1 & k=1 \
0 & textelse,
endcases
$$
the same as $e_2 wedge cdots wedge e_n$. Finally, let $omega in A_m-1(S)$ be arbitrary. By surjectivity of $phi$, we find $v in S$ so that $phi(v) = omega$. Then $phi(cv) = comega$, where $c$ is a constant such that $|cv| = 1$. Then extend $v$ to an orthonormal basis by basis extension and Gram‒Schmidt and apply the second point to get that $comega$ is decomposable. Hence $omega$ is decomposable by sliding in the $c$ into one of the wedge factors by multilinearity.
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your idea is indeed a correct proof; maybe in the third equivalent statement in the highlighted row one could insert "for all $v_1, ldots, v_m-1$".
For the second statement, it suffices to show that the two $m-1$-forms coincide on the given orthonormal positive basis, which we denote by $v = b_1, b_2, ldots, b_n$; but indeed,
$$
phi(v)(b_1, ldots, hatb_k, ldots, b_n) = langle b_1, b_1, ldots, hatb_k, ldots, b_n rangle = begincases
+1 & k=1 \
0 & textelse,
endcases
$$
the same as $e_2 wedge cdots wedge e_n$. Finally, let $omega in A_m-1(S)$ be arbitrary. By surjectivity of $phi$, we find $v in S$ so that $phi(v) = omega$. Then $phi(cv) = comega$, where $c$ is a constant such that $|cv| = 1$. Then extend $v$ to an orthonormal basis by basis extension and Gram‒Schmidt and apply the second point to get that $comega$ is decomposable. Hence $omega$ is decomposable by sliding in the $c$ into one of the wedge factors by multilinearity.
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
add a comment |Â
up vote
0
down vote
Your idea is indeed a correct proof; maybe in the third equivalent statement in the highlighted row one could insert "for all $v_1, ldots, v_m-1$".
For the second statement, it suffices to show that the two $m-1$-forms coincide on the given orthonormal positive basis, which we denote by $v = b_1, b_2, ldots, b_n$; but indeed,
$$
phi(v)(b_1, ldots, hatb_k, ldots, b_n) = langle b_1, b_1, ldots, hatb_k, ldots, b_n rangle = begincases
+1 & k=1 \
0 & textelse,
endcases
$$
the same as $e_2 wedge cdots wedge e_n$. Finally, let $omega in A_m-1(S)$ be arbitrary. By surjectivity of $phi$, we find $v in S$ so that $phi(v) = omega$. Then $phi(cv) = comega$, where $c$ is a constant such that $|cv| = 1$. Then extend $v$ to an orthonormal basis by basis extension and Gram‒Schmidt and apply the second point to get that $comega$ is decomposable. Hence $omega$ is decomposable by sliding in the $c$ into one of the wedge factors by multilinearity.
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your idea is indeed a correct proof; maybe in the third equivalent statement in the highlighted row one could insert "for all $v_1, ldots, v_m-1$".
For the second statement, it suffices to show that the two $m-1$-forms coincide on the given orthonormal positive basis, which we denote by $v = b_1, b_2, ldots, b_n$; but indeed,
$$
phi(v)(b_1, ldots, hatb_k, ldots, b_n) = langle b_1, b_1, ldots, hatb_k, ldots, b_n rangle = begincases
+1 & k=1 \
0 & textelse,
endcases
$$
the same as $e_2 wedge cdots wedge e_n$. Finally, let $omega in A_m-1(S)$ be arbitrary. By surjectivity of $phi$, we find $v in S$ so that $phi(v) = omega$. Then $phi(cv) = comega$, where $c$ is a constant such that $|cv| = 1$. Then extend $v$ to an orthonormal basis by basis extension and Gram‒Schmidt and apply the second point to get that $comega$ is decomposable. Hence $omega$ is decomposable by sliding in the $c$ into one of the wedge factors by multilinearity.
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
Your idea is indeed a correct proof; maybe in the third equivalent statement in the highlighted row one could insert "for all $v_1, ldots, v_m-1$".
For the second statement, it suffices to show that the two $m-1$-forms coincide on the given orthonormal positive basis, which we denote by $v = b_1, b_2, ldots, b_n$; but indeed,
$$
phi(v)(b_1, ldots, hatb_k, ldots, b_n) = langle b_1, b_1, ldots, hatb_k, ldots, b_n rangle = begincases
+1 & k=1 \
0 & textelse,
endcases
$$
the same as $e_2 wedge cdots wedge e_n$. Finally, let $omega in A_m-1(S)$ be arbitrary. By surjectivity of $phi$, we find $v in S$ so that $phi(v) = omega$. Then $phi(cv) = comega$, where $c$ is a constant such that $|cv| = 1$. Then extend $v$ to an orthonormal basis by basis extension and Gram‒Schmidt and apply the second point to get that $comega$ is decomposable. Hence $omega$ is decomposable by sliding in the $c$ into one of the wedge factors by multilinearity.
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
answered Jul 17 at 6:56
AlgebraicsAnonymous
69111
69111
add a comment |Â
add a comment |Â
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