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What is the difference between differential 1-form and duality pairing?
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I am trying to learn the basics of differential geometry. The duality pairing between a coordinate basis and its canonical basis is given by the Kronecker delta, which is nothing but a scalar value. Whereas, the differential 1-form in my notes is given as $alpha = alpha_i (x^1,...,x^n) dx^i$. In the above relation, am i right in assuming that $(x^1,...,x^n)$ is the vector of expansion coefficients of any tangent vector, if $alpha_i$ are the expansion coefficients of the corresponding cotangent vector? If so, how does the relation for differential 1-form has a dimension of 1? Also how does one physically interpret duality pairing and differential 1-form?
general-topology differential-geometry
asked Jul 22 at 7:59
K. Sripathy
31
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up vote
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I am trying to learn the basics of differential geometry. The duality pairing between a coordinate basis and its canonical basis is given by the Kronecker delta, which is nothing but a scalar value. Whereas, the differential 1-form in my notes is given as $alpha = alpha_i (x^1,...,x^n) dx^i$. In the above relation, am i right in assuming that $(x^1,...,x^n)$ is the vector of expansion coefficients of any tangent vector, if $alpha_i$ are the expansion coefficients of the corresponding cotangent vector? If so, how does the relation for differential 1-form has a dimension of 1? Also how does one physically interpret duality pairing and differential 1-form?
general-topology differential-geometry
asked Jul 22 at 7:59
K. Sripathy
31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to learn the basics of differential geometry. The duality pairing between a coordinate basis and its canonical basis is given by the Kronecker delta, which is nothing but a scalar value. Whereas, the differential 1-form in my notes is given as $alpha = alpha_i (x^1,...,x^n) dx^i$. In the above relation, am i right in assuming that $(x^1,...,x^n)$ is the vector of expansion coefficients of any tangent vector, if $alpha_i$ are the expansion coefficients of the corresponding cotangent vector? If so, how does the relation for differential 1-form has a dimension of 1? Also how does one physically interpret duality pairing and differential 1-form?
general-topology differential-geometry
asked Jul 22 at 7:59
K. Sripathy
31
I am trying to learn the basics of differential geometry. The duality pairing between a coordinate basis and its canonical basis is given by the Kronecker delta, which is nothing but a scalar value. Whereas, the differential 1-form in my notes is given as $alpha = alpha_i (x^1,...,x^n) dx^i$. In the above relation, am i right in assuming that $(x^1,...,x^n)$ is the vector of expansion coefficients of any tangent vector, if $alpha_i$ are the expansion coefficients of the corresponding cotangent vector? If so, how does the relation for differential 1-form has a dimension of 1? Also how does one physically interpret duality pairing and differential 1-form?
general-topology differential-geometry
asked Jul 22 at 7:59
K. Sripathy
31
asked Jul 22 at 7:59
K. Sripathy
31
asked Jul 22 at 7:59
K. Sripathy
31
asked Jul 22 at 7:59
K. Sripathy
31
31
add a comment |Â
add a comment |Â
1 Answer
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oldest
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up vote
0
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To fix the language: For a finite dimensional real vector space $V$, its dual space is defined as
$$V^* := alpha: V to mathbb R mid alpha text is linear .$$
A dual pairing is now a map
$$<cdot,cdot>: V times V^* to mathbb R: (v,alpha) mapsto alpha(v). $$
If $ e_1, dots, e_n$ is a basis of $V$, we call $ e^1, dots, e^n$ the dual basis of $V^*$, if
$$ e^i(e_j) = delta_ij, text for all i,j.$$
Is $alpha(x^1, dots, x^n)$ the vector of expansion coefficients of any tangent vector?
No. I would assume, that these are coefficients which depend on the current coordinates.
A differential 1-form is a smooth map $alpha:TM to mathbb R$ which is linear on each fiber, i.e. at each point $p$, the restriction $alpha_p: T_p M to mathbb R$ is a linear map, or equivalently $alpha_p in (T_pM)^* =: T^*_pM$
A given chart implies a basis of the tangent space $left fracpartialpartial x^1 subset T_pM$ and a dual basis of the cotangent space $left dx^1_p, dots, dx^n_p right subset T^*_p M$.
Now at each point $p$ we can find coefficients $lambda_i(p)$, to write the covector $alpha_p$ as a linear combination of the dual basis vectors, i.e. $alpha_p = sum_i lambda_i(p) dx^i_p.$
But these coefficients depend on the current point $p$!
Therefore, if we want to describe $alpha$ not only at $p$ but in the domain of a chart $x$, we may write
$$ alpha = sum_i lambda_i(x^1, dots, x^n) dx^i.$$
Relation to the dual pairing?
The dual pairing is already hidden in the construction of the dual basis!
For tangential and cotangential vectors, the relation between a basis and a dual basis reads
$$ dx^jleft( fracpartialpartial x^i right) = delta_ij.$$
In particular
$$alphaleft(fracpartialpartial x^jright) = sum_i lambda_i dx^ileft(fracpartialpartial x^jright) = lambda_j.$$
Physical interpretation
That's a question on its own. There are many important 1-forms in physics, for example momentum and (often) force. In physics these are usually called contravariant vectors, since they transform differently than tangent vectors.
answered Jul 22 at 11:33
Steffen Plunder
45829
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
To fix the language: For a finite dimensional real vector space $V$, its dual space is defined as
$$V^* := alpha: V to mathbb R mid alpha text is linear .$$
A dual pairing is now a map
$$<cdot,cdot>: V times V^* to mathbb R: (v,alpha) mapsto alpha(v). $$
If $ e_1, dots, e_n$ is a basis of $V$, we call $ e^1, dots, e^n$ the dual basis of $V^*$, if
$$ e^i(e_j) = delta_ij, text for all i,j.$$
Is $alpha(x^1, dots, x^n)$ the vector of expansion coefficients of any tangent vector?
No. I would assume, that these are coefficients which depend on the current coordinates.
A differential 1-form is a smooth map $alpha:TM to mathbb R$ which is linear on each fiber, i.e. at each point $p$, the restriction $alpha_p: T_p M to mathbb R$ is a linear map, or equivalently $alpha_p in (T_pM)^* =: T^*_pM$
A given chart implies a basis of the tangent space $left fracpartialpartial x^1 subset T_pM$ and a dual basis of the cotangent space $left dx^1_p, dots, dx^n_p right subset T^*_p M$.
Now at each point $p$ we can find coefficients $lambda_i(p)$, to write the covector $alpha_p$ as a linear combination of the dual basis vectors, i.e. $alpha_p = sum_i lambda_i(p) dx^i_p.$
But these coefficients depend on the current point $p$!
Therefore, if we want to describe $alpha$ not only at $p$ but in the domain of a chart $x$, we may write
$$ alpha = sum_i lambda_i(x^1, dots, x^n) dx^i.$$
Relation to the dual pairing?
The dual pairing is already hidden in the construction of the dual basis!
For tangential and cotangential vectors, the relation between a basis and a dual basis reads
$$ dx^jleft( fracpartialpartial x^i right) = delta_ij.$$
In particular
$$alphaleft(fracpartialpartial x^jright) = sum_i lambda_i dx^ileft(fracpartialpartial x^jright) = lambda_j.$$
Physical interpretation
That's a question on its own. There are many important 1-forms in physics, for example momentum and (often) force. In physics these are usually called contravariant vectors, since they transform differently than tangent vectors.
answered Jul 22 at 11:33
Steffen Plunder
45829
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
add a comment |Â
up vote
0
down vote
accepted
To fix the language: For a finite dimensional real vector space $V$, its dual space is defined as
$$V^* := alpha: V to mathbb R mid alpha text is linear .$$
A dual pairing is now a map
$$<cdot,cdot>: V times V^* to mathbb R: (v,alpha) mapsto alpha(v). $$
If $ e_1, dots, e_n$ is a basis of $V$, we call $ e^1, dots, e^n$ the dual basis of $V^*$, if
$$ e^i(e_j) = delta_ij, text for all i,j.$$
Is $alpha(x^1, dots, x^n)$ the vector of expansion coefficients of any tangent vector?
No. I would assume, that these are coefficients which depend on the current coordinates.
A differential 1-form is a smooth map $alpha:TM to mathbb R$ which is linear on each fiber, i.e. at each point $p$, the restriction $alpha_p: T_p M to mathbb R$ is a linear map, or equivalently $alpha_p in (T_pM)^* =: T^*_pM$
A given chart implies a basis of the tangent space $left fracpartialpartial x^1 subset T_pM$ and a dual basis of the cotangent space $left dx^1_p, dots, dx^n_p right subset T^*_p M$.
Now at each point $p$ we can find coefficients $lambda_i(p)$, to write the covector $alpha_p$ as a linear combination of the dual basis vectors, i.e. $alpha_p = sum_i lambda_i(p) dx^i_p.$
But these coefficients depend on the current point $p$!
Therefore, if we want to describe $alpha$ not only at $p$ but in the domain of a chart $x$, we may write
$$ alpha = sum_i lambda_i(x^1, dots, x^n) dx^i.$$
Relation to the dual pairing?
The dual pairing is already hidden in the construction of the dual basis!
For tangential and cotangential vectors, the relation between a basis and a dual basis reads
$$ dx^jleft( fracpartialpartial x^i right) = delta_ij.$$
In particular
$$alphaleft(fracpartialpartial x^jright) = sum_i lambda_i dx^ileft(fracpartialpartial x^jright) = lambda_j.$$
Physical interpretation
That's a question on its own. There are many important 1-forms in physics, for example momentum and (often) force. In physics these are usually called contravariant vectors, since they transform differently than tangent vectors.
answered Jul 22 at 11:33
Steffen Plunder
45829
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
To fix the language: For a finite dimensional real vector space $V$, its dual space is defined as
$$V^* := alpha: V to mathbb R mid alpha text is linear .$$
A dual pairing is now a map
$$<cdot,cdot>: V times V^* to mathbb R: (v,alpha) mapsto alpha(v). $$
If $ e_1, dots, e_n$ is a basis of $V$, we call $ e^1, dots, e^n$ the dual basis of $V^*$, if
$$ e^i(e_j) = delta_ij, text for all i,j.$$
Is $alpha(x^1, dots, x^n)$ the vector of expansion coefficients of any tangent vector?
No. I would assume, that these are coefficients which depend on the current coordinates.
A differential 1-form is a smooth map $alpha:TM to mathbb R$ which is linear on each fiber, i.e. at each point $p$, the restriction $alpha_p: T_p M to mathbb R$ is a linear map, or equivalently $alpha_p in (T_pM)^* =: T^*_pM$
A given chart implies a basis of the tangent space $left fracpartialpartial x^1 subset T_pM$ and a dual basis of the cotangent space $left dx^1_p, dots, dx^n_p right subset T^*_p M$.
Now at each point $p$ we can find coefficients $lambda_i(p)$, to write the covector $alpha_p$ as a linear combination of the dual basis vectors, i.e. $alpha_p = sum_i lambda_i(p) dx^i_p.$
But these coefficients depend on the current point $p$!
Therefore, if we want to describe $alpha$ not only at $p$ but in the domain of a chart $x$, we may write
$$ alpha = sum_i lambda_i(x^1, dots, x^n) dx^i.$$
Relation to the dual pairing?
The dual pairing is already hidden in the construction of the dual basis!
For tangential and cotangential vectors, the relation between a basis and a dual basis reads
$$ dx^jleft( fracpartialpartial x^i right) = delta_ij.$$
In particular
$$alphaleft(fracpartialpartial x^jright) = sum_i lambda_i dx^ileft(fracpartialpartial x^jright) = lambda_j.$$
Physical interpretation
That's a question on its own. There are many important 1-forms in physics, for example momentum and (often) force. In physics these are usually called contravariant vectors, since they transform differently than tangent vectors.
answered Jul 22 at 11:33
Steffen Plunder
45829
To fix the language: For a finite dimensional real vector space $V$, its dual space is defined as
$$V^* := alpha: V to mathbb R mid alpha text is linear .$$
A dual pairing is now a map
$$<cdot,cdot>: V times V^* to mathbb R: (v,alpha) mapsto alpha(v). $$
If $ e_1, dots, e_n$ is a basis of $V$, we call $ e^1, dots, e^n$ the dual basis of $V^*$, if
$$ e^i(e_j) = delta_ij, text for all i,j.$$
Is $alpha(x^1, dots, x^n)$ the vector of expansion coefficients of any tangent vector?
No. I would assume, that these are coefficients which depend on the current coordinates.
A differential 1-form is a smooth map $alpha:TM to mathbb R$ which is linear on each fiber, i.e. at each point $p$, the restriction $alpha_p: T_p M to mathbb R$ is a linear map, or equivalently $alpha_p in (T_pM)^* =: T^*_pM$
A given chart implies a basis of the tangent space $left fracpartialpartial x^1 subset T_pM$ and a dual basis of the cotangent space $left dx^1_p, dots, dx^n_p right subset T^*_p M$.
Now at each point $p$ we can find coefficients $lambda_i(p)$, to write the covector $alpha_p$ as a linear combination of the dual basis vectors, i.e. $alpha_p = sum_i lambda_i(p) dx^i_p.$
But these coefficients depend on the current point $p$!
Therefore, if we want to describe $alpha$ not only at $p$ but in the domain of a chart $x$, we may write
$$ alpha = sum_i lambda_i(x^1, dots, x^n) dx^i.$$
Relation to the dual pairing?
The dual pairing is already hidden in the construction of the dual basis!
For tangential and cotangential vectors, the relation between a basis and a dual basis reads
$$ dx^jleft( fracpartialpartial x^i right) = delta_ij.$$
In particular
$$alphaleft(fracpartialpartial x^jright) = sum_i lambda_i dx^ileft(fracpartialpartial x^jright) = lambda_j.$$
Physical interpretation
That's a question on its own. There are many important 1-forms in physics, for example momentum and (often) force. In physics these are usually called contravariant vectors, since they transform differently than tangent vectors.
answered Jul 22 at 11:33
Steffen Plunder
45829
answered Jul 22 at 11:33
Steffen Plunder
45829
answered Jul 22 at 11:33
Steffen Plunder
45829
answered Jul 22 at 11:33
Steffen Plunder
45829
45829
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
add a comment |Â
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
Thanks for the clarification. It makes more sense now that the expansion coefficient of co-vector is a function of the position in tangential vector space. I tend to get lost in the syntax for these expressions. Hypothetically, had it been a vector of expansion coefficients, would I represent it with $x^j$ in that case?
– K. Sripathy
Jul 22 at 11:47
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
If you have a tangent vector $v in TM$, you could represent it with respect to a chart $x$ via $v = sum mu_i fracpartialpartial x^i = sum_i dx^i( v ) fracpartialpartial x^i$. A bit confusing, I know... So, if you want to get a (locally defined) chart $TM to mathbb R^2n$, it would be $mapsto left(x^1(p), dots x^n(p), dx^1(v), dots dx^n(v)right)$. Here $p$ is the point such that $v in T_pM$. The expansion coefficients of $v$ with respect the chart's tangent basis are $dx^i(v)$.
– Steffen Plunder
Jul 22 at 11:59
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
Much appreciated!
– K. Sripathy
Jul 22 at 12:43
add a comment |Â
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