Differential Forms: Definition & Antisymmetric Tensor

[kent davidge]

Why does the definition of a differential form requires a totally antisymmetric tensor?

 

[Orodruin]

Why does the definition of a differential form requires a totally antisymmetric tensor?

Because that is how we define a differential form. It is a definition that turns out to be useful in describing several things, but ultimately it is just that - a definition.

 

[fresh_42]

Why does the definition of a differential form requires a totally antisymmetric tensor?

It doesn't require tensors.
It requires multi-linearity (linear in each argument) and alternating. The latter is equivalent to $d(X,X)=0$.

Tensors are useful, when you try to make sure, that the definition of a multi-linear object actually defines something, i.e. something that exists, rather than being a funny definition, that describes the empty set.
The reason for this is algebraic and somehow abstract, so I leave it here in the relativity section. For short: tensors are a kind of a universal multi-linear object, and special multi-linear objects like differential forms can be seen as the image under a certain projection. Even shorter: Tensors are a skeleton key.

 

[robphy]

Possibly useful reading:
http://mathoverflow.net/questions/25089/basic-question-about-differential-forms-and-physics

https://en.wikipedia.org/wiki/Hermann_Grassmann#Mathematician
http://www-history.mcs.st-and.ac.uk/Extras/Grassmann_1844.html from Grassmann's 1844 "The Theory of Linear Extension, a New Branch of Mathematics"

https://en.wikipedia.org/wiki/Exterior_algebra

http://bookstore.ams.org/hmath-19/ Grassmann's 1862 "Extension Theory"

 

[vanhees71]

It doesn't require tensors.
It requires multi-linearity (linear in each argument) and alternating. The latter is equivalent to $d(X,X)=0$.

Tensors are useful, when you try to make sure, that the definition of a multi-linear object actually defines something, i.e. something that exists, rather than being a funny definition, that describes the empty set.
The reason for this is algebraic and somehow abstract, so I leave it here in the relativity section. For short: tensors are a kind of a universal multi-linear object, and special multi-linear objects like differential forms can be seen as the image under a certain projection. Even shorter: Tensors are a skeleton key.

Well tensors are defined as multilinear forms (not necessarily antisymmetric ones as the alternating forms in Cartan calculus), i.e., a tensor of rank $n$ is a linear map from $V^n$ into the field of scalars of $V$ (field meant here in the algebraic sense).

 

[kent davidge]

Thanks for answering. Why integration and differentiation in general is chosen to be performed with antisymmetric(0,n) tensors ? Why not on other type of tensors?

 

[Orodruin]

Thanks for answering. Why integration and differentiation in general is chosen to be performed with antisymmetric(0,n) tensors ? Why not on other type of tensors?

Because a directed volume element spanned by n vectors is naturally described by an antisymmetric tensor of rank n.

 

[kent davidge]

Because a directed volume element spanned by n vectors is naturally described by an antisymmetric tensor of rank n.

Could you give me a simple (maybe a 3-dimensional) example?

 

[vanhees71]

Take a surface $\vec{x}=\vec{x}(u,v)$ in $\mathbb{R}^3$. Then the surface-normal vector is determined by the parallelogram defined by the infinitesimal pieces of the coordinate lines $v=\text{const}$ and $u=\text{const}$. The magnitude and direction is given by the cross product, i.e., you have
$$\mathrm{d}^2 \vec{a}=\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v}.$$
Note that the orientation, i.e., the direction of the normal vector depends in which order you choose the indepenent variables $u$ and $v$ (i.e., you have defined an oriented surface).

In an arbitrary (pseudo-)Riemannian manifold hypersurface elements are defined in the analogous general covariant way with help of the Levi-Civita tensor,
$$\epsilon^{\mu_1 \mu_2 \ldots \mu_d}=\sqrt{|g|} \Delta^{\mu_1 \ldots \mu_d},$$
where $\Delta^{\mu_1 \ldots \mu_d}$ is the totally antisymmetric Levi-Civita symbol with $\Delta^{12\ldots d}=1$ and $g=\mathrm{det} (g_{\mu \nu})$ the determinant of the (pseudo-)metric components. A $k$-dimensional ($k \leq d$) hyper-surface element is defined by some function $q^{\mu}(u_1,\ldots,u_k)$ and the corresponding integral measure by
$$\mathrm{d}^k \Sigma^{\mu_{k+1}\ldots \mu_d}=\mathrm{d} q_1 \cdots \mathrm{d} q_d \epsilon_{\mu_1\ldots \mu_d} \frac{\partial q^{\mu_1}}{\partial u_1} \cdots \frac{\partial q^{\mu_k}}{\partial u_k}.$$

 

[robphy]

Could you give me a simple (maybe a 3-dimensional) example?

Here's a 2-D example... a parallelogram formed from two vectors.
If you specify an ordered-pair with those two vectors, then you have specified an oriented parallelogram.
The size of the parallelogram can be expressed using the determinant formed using the components of the vectors [although it could be expressed in a coordinate-free way].

 

[kent davidge]

Take a surface $\vec{x}=\vec{x}(u,v)$ in $\mathbb{R}^3$. Then the surface-normal vector is determined by the parallelogram defined by the infinitesimal pieces of the coordinate lines $v=\text{const}$ and $u=\text{const}$. The magnitude and direction is given by the cross product, i.e., you have
$$\mathrm{d}^2 \vec{a}=\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v}.$$
Note that the orientation, i.e., the direction of the normal vector depends in which order you choose the indepenent variables $u$ and $v$ (i.e., you have defined an oriented surface).

In an arbitrary (pseudo-)Riemannian manifold hypersurface elements are defined in the analogous general covariant way with help of the Levi-Civita tensor,
$$\epsilon^{\mu_1 \mu_2 \ldots \mu_d}=\sqrt{|g|} \Delta^{\mu_1 \ldots \mu_d},$$
where $\Delta^{\mu_1 \ldots \mu_d}$ is the totally antisymmetric Levi-Civita symbol with $\Delta^{12\ldots d}=1$ and $g=\mathrm{det} (g_{\mu \nu})$ the determinant of the (pseudo-)metric components. A $k$-dimensional ($k \leq d$) hyper-surface element is defined by some function $q^{\mu}(u_1,\ldots,u_k)$ and the corresponding integral measure by
$$\mathrm{d}^k \Sigma^{\mu_{k+1}\ldots \mu_d}=\mathrm{d} q_1 \cdots \mathrm{d} q_d \epsilon_{\mu_1\ldots \mu_d} \frac{\partial q^{\mu_1}}{\partial u_1} \cdots \frac{\partial q^{\mu_k}}{\partial u_k}.$$

In your first example in R³, $$\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial u} \times \frac{\partial \vec{x}}{\partial v}= -\mathrm{d} u \mathrm{d} v \frac{\partial \vec{x}}{\partial v} \times \frac{\partial \vec{x}}{\partial u}.$$ Right? So is it a antisymmetric tensor?

Here's a 2-D example... a parallelogram formed from two vectors.
If you specify an ordered-pair with those two vectors, then you have specified an oriented parallelogram.
The size of the parallelogram can be expressed using the determinant formed using the components of the vectors [although it could be expressed in a coordinate-free way].

Where does the antisymmetry property comes in this situation?

(I'm sorry for my bad english)

 

[robphy]

Where does the antisymmetry property comes in this situation?

If you switch the order of two vectors, it is multiplied by a minus sign... and this shows up in the determinant.

 

[Orodruin]

So is it a antisymmetric tensor?

Yes, it is a contraction between the two tangent vectors $\partial\vec x/\partial u$ and $\partial\vec x/\partial v$ with the anti-symmetric tensor $\epsilon_{ijk}$.

 

[kent davidge]

Thanks!

 

[stevendaryl]

Thanks for answering. Why integration and differentiation in general is chosen to be performed with antisymmetric(0,n) tensors ? Why not on other type of tensors?

The simplest case to understand is integration over a curved 2-D manifold with a metric. And let me assume a Riemannian metric (so we can assume that the square of any vector is defined and positive). The case for an indefinite metric (so the square of a vector can be negative or zero) and for higher-dimensional manifolds will be left as an exercise .

Using the usual definition of an integral, if we are integrating a scalar function $F$ over the 2-D space, we can approximate it by dividing space up into lots of little parallelograms, and compute

$\int F = \sum_i \mathcal{A}_i F(\mathcal{P}_i)$

where $\mathcal{A}_i$ is the area of parallelogram number $i$, and $F(\mathcal{P}_i)$ is the value of $F$ at the point $\mathcal{P}_i$ somewhere inside the parallelogram. So integrating a scalar function over a 2-D manifold is reduced to the problem of computing the area of a tiny parallelogram.

The rule for the area of a parallelogram in good old Euclidean geometry is just:

$\mathcal{A} = base \times height$

So if $A$ is the displacement vector (which is approximately well-defined for small displacements, even in curved space) for one side of a small parallelogram, and $B$ is the displacement vector for the other side, then the area will be:

$\mathcal{A} = |A| |B^\perp|$

where $B^\perp$ is the component of $B$ perpendicular to $A$. (And where $|X|$ means the square-root of $X \cdot X$, which is real and positive for Riemannian manifolds, which we're assuming.)

We can compute $|B^\perp|$ in terms of the vector dot-product. $|B^\perp| = \sqrt{B \cdot B - \frac{(A \cdot B)^2}{A \cdot A}}$. So we have:

$\mathcal{A} = |A| \sqrt{B \cdot B - \frac{(A \cdot B)^2}{A \cdot A}} = \sqrt{(A \cdot A) (B \cdot B) - (A \cdot B)^2}$

Or squaring both sides gives:
$\mathcal{A}^2 = (A \cdot A) (B \cdot B) - (A \cdot B)^2$

At this point, let me introduce the alternative way of writing vector dot products. The dot product $X \cdot Y$ can be written in terms of the metric tensor $g_{\mu \nu}$ as $X \cdot Y = \sum_{\mu \nu} g_{\mu \nu} X^\mu Y^\nu \equiv X^\mu Y_\mu$, where $Y_\mu \equiv \sum_{\nu} g_{\mu \nu} Y^\nu$. So in this notation, we can write:

$\mathcal{A}^2 = A^\mu A_\mu B^\alpha B_\alpha - A^\mu B_\mu A^\alpha B_\alpha$

Now, introduce the two-index tensor $\mathcal{A}^{\mu \alpha} \equiv A^\mu B^\alpha - A^\alpha B^\mu$. We can prove the following identity:

$\mathcal{A}^{\mu \alpha} \mathcal{A}_{\mu \alpha} = (A^\mu B^\alpha - A^\alpha B^\mu)(A_\mu B_\alpha - A_\alpha B_\mu) = A^\mu A_\mu B^\alpha B_\alpha - A^\mu B_\mu A^\alpha B_\alpha - A^\alpha B_\alpha A^\mu B_\mu + A^\alpha A_\alpha B^\mu B_\mu = (A \cdot A) (B \cdot B) - (A \cdot B) (A \cdot B) - (A \cdot B) (A \cdot B) + (A \cdot A) (B \cdot B) = 2 ((A \cdot A) (B \cdot B) - (A \cdot B) (A \cdot B)) = 2 \mathcal{A}^2$
So we have:
$\mathcal{A}^2 = \frac{1}{2}\mathcal{A}^{\mu \alpha} \mathcal{A}_{\mu \alpha}$. (The factor of 1/2 is a little worrisome, but I think it's because the implicit summing over $\mu, \nu, \alpha, \beta$ double-counts; the sum includes the non-summed term $A^\mu A_\mu B^\alpha B_\alpha$ and $A^\alpha A_\alpha B^\mu B_\mu$ for each $\alpha$ and $\mu$.)

So the area tensor for the parallelogram of vectors $A$ and $B$ is antisymmetric.

 

[kent davidge]

The simplest case to understand is integration over a curved 2-D manifold with a metric. And let me assume a Riemannian metric (so we can assume that the square of any vector is defined and positive). The case for an indefinite metric (so the square of a vector can be negative or zero) and for higher-dimensional manifolds will be left as an exercise .

Using the usual definition of an integral, if we are integrating a scalar function $F$ over the 2-D space, we can approximate it by dividing space up into lots of little parallelograms, and compute

$\int F = \sum_i \mathcal{A}_i F(\mathcal{P}_i)$

View attachment 107924

where $\mathcal{A}_i$ is the area of parallelogram number $i$, and $F(\mathcal{P}_i)$ is the value of $F$ at the point $\mathcal{P}_i$ somewhere inside the parallelogram. So integrating a scalar function over a 2-D manifold is reduced to the problem of computing the area of a tiny parallelogram.

The rule for the area of a parallelogram in good old Euclidean geometry is just:

$\mathcal{A} = base \times height$

So if $A$ is the displacement vector (which is approximately well-defined for small displacements, even in curved space) for one side of a small parallelogram, and $B$ is the displacement vector for the other side, then the area will be:
View attachment 107925
$\mathcal{A} = |A| |B^\perp|$

where $B^\perp$ is the component of $B$ perpendicular to $A$. (And where $|X|$ means the square-root of $X \cdot X$, which is real and positive for Riemannian manifolds, which we're assuming.)

We can compute $|B^\perp|$ in terms of the vector dot-product. $|B^\perp| = \sqrt{B \cdot B - \frac{(A \cdot B)^2}{A \cdot A}}$. So we have:

$\mathcal{A} = |A| \sqrt{B \cdot B - \frac{(A \cdot B)^2}{A \cdot A}} = \sqrt{(A \cdot A) (B \cdot B) - (A \cdot B)^2}$

Or squaring both sides gives:
$\mathcal{A}^2 = (A \cdot A) (B \cdot B) - (A \cdot B)^2$

At this point, let me introduce the alternative way of writing vector dot products. The dot product $X \cdot Y$ can be written in terms of the metric tensor $g_{\mu \nu}$ as $X \cdot Y = \sum_{\mu \nu} g_{\mu \nu} X^\mu Y^\nu \equiv X^\mu Y_\mu$, where $Y_\mu \equiv \sum_{\nu} g_{\mu \nu} Y^\nu$. So in this notation, we can write:

$\mathcal{A}^2 = A^\mu A_\mu B^\alpha B_\alpha - A^\mu B_\mu A^\alpha B_\alpha$

Now, introduce the two-index tensor $\mathcal{A}^{\mu \alpha} \equiv A^\mu B^\alpha - A^\alpha B^\mu$. We can prove the following identity:

$\mathcal{A}^{\mu \alpha} \mathcal{A}_{\mu \alpha} = (A^\mu B^\alpha - A^\alpha B^\mu)(A_\mu B_\alpha - A_\alpha B_\mu) = A^\mu A_\mu B^\alpha B_\alpha - A^\mu B_\mu A^\alpha B_\alpha - A^\alpha B_\alpha A^\mu B_\mu + A^\alpha A_\alpha B^\mu B_\mu = (A \cdot A) (B \cdot B) - (A \cdot B) (A \cdot B) - (A \cdot B) (A \cdot B) + (A \cdot A) (B \cdot B) = 2 ((A \cdot A) (B \cdot B) - (A \cdot B) (A \cdot B)) = 2 \mathcal{A}^2$
So we have:
$\mathcal{A}^2 = \frac{1}{2}\mathcal{A}^{\mu \alpha} \mathcal{A}_{\mu \alpha}$. (The factor of 1/2 is a little worrisome, but I think it's because the implicit summing over $\mu, \nu, \alpha, \beta$ double-counts; the sum includes the non-summed term $A^\mu A_\mu B^\alpha B_\alpha$ and $A^\alpha A_\alpha B^\mu B_\mu$ for each $\alpha$ and $\mu$.)

So the area tensor for the parallelogram of vectors $A$ and $B$ is antisymmetric.

Thank you for your detailed explanation. So the conclusion is that $\mathcal{A}_{\mu \alpha}$ must be antisymmetric, since A² is (in this case) always positive, so that if we change μ and α, we have ($\mathcal{A}^{\alpha \mu}$)($\mathcal{A}_{\alpha \mu})=$ $(\mathcal{-A}^{\mu \alpha})$$(\mathcal{-A}_{\mu \alpha})$ in order to get the same value for A². Am I right?