Technical question about 2-forms

Question

A technical question about the electromagnetic tensor, but before that, it is know if, say, instead of being $$F_{\mu\nu}=\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$it were $$F_{\mu\nu}=(...)_{\mu\nu}-(....)_{\mu\nu}$$ I am giving an example of the electromagnetric tensor because it is a 2-form, instead of F you can tay any 2-form.

Now my question is, can this be written (As a short cut or whatever) as $$F=(...)-(....)$$ with swallowing the indices?

This procedure I saw in some 1-forms, of only one index ($\mu$) $$A_\mu=(...)_{\mu}-(....)_{\mu}$$ and it then is the case where we can swallow the index and we get $$A=(...)-(....)$$ I was wondering if we can do that in the 2-form case and why is it applicable?

Answer 1

We do not "swallow" the index. You must distignuish between the geometrical object and its components, and that is not unique to forms, but occurs for all vectors:

If you have a vector $v\in V$, where $V$ is some vector space, it has no indices. It's just an element. Now, if you choose a basis $e_1,\dots,e_n$ of the space, you write $$ v = v^\mu e_\mu$$ and physicists often call the collection of components $v^\mu$ together with their transformation behaviour "the vector".

Now, a $k$-form $\omega$ on a (spacetime) manifold $M$ is a map that assigns to every point $p\in M$ an antisymmetric $k$-tensor $\omega_p$, i.e. $\omega_p$ eats $k$ tangent vectors $v_1,\dots,v_n\in T_p M$ (formally, we have $\omega_p \in \bigwedge T^*_p M$). This means $\omega_p(v_1,\dots,v_n)$ is finally a number, and, as a geometric object, $\omega$ is a map $M \to \bigwedge^k T^* M$.

Now, physicists don't like to talk about these abstract objects for some reason, and prefer dealing with their components alone: Given coordinates $x^\mu$ on $M$, the tangent vectors have a natural basis in the $\partial_\mu = \frac{\partial}{\partial x^\mu}$, i.e. we can expand $v = v^\mu \partial_\mu$.

Correspondingly, the cotangent space $T_p^* M$ has a dual basis denoted by $\mathrm{d}x^\mu$ defined by the relation $\mathrm{d}x^\mu(\partial_\nu) = \delta^\mu{}_\nu$.

We now just write $x$ for a point on the manifold and have now two ways to look at the "components" $\omega_{\mu_1\dots\mu_k}(x)$ of $\omega$:

• Put the basis vectors in: $\omega_{\mu_1\dots\mu_k}(x) := \omega_x(\partial_{\mu_1},\dots,\partial_{\mu_k})$

• Expand $\omega_x$ in the basis $\mathrm{d}x^{\mu_1}\wedge\dots\wedge\mathrm{d}x^{\mu_k}$ of wedges of cotangent vectors: $$ \omega_x = c_{\mu_1\dots\mu_k}\mathrm{d}x^{\mu_1}\wedge\dots\wedge\mathrm{d}x^{\mu_k}$$ and call the coefficients $c_{\mu_1\dots\mu_k}$ then $\omega_{\mu_1\dots\mu_k}(x)$.

So, when a physicist writes $F^{\mu\nu}$, they mean by definition that there is a $2$ form $$ F : M \to T^* M \wedge T^* M$$ which may be written as $$ F = F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu$$ or, depending on the convention, as $$ F = \frac{1}{2}F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu$$ since many people introduce a $\frac{1}{k!}$ into the components of a $k$-form to cancel out the "overcounting" of values that occurs due to the antisymmetry properties.

To finally answer your question:

No, $F$ is not a shorthand for $F^{\mu\nu}$ in the usual sense of a short hand. $F$ is the actual geometric object and $F^{\mu\nu}$ are its components in a particular coordinate basis. And, whatever you may have seen, we have $A = A_\mu\mathrm{d}x^\mu$ for $1$-forms, so you may also not use $A$ as a "shorthand" for $A^\mu$ - again, one is the geometric object, the other are its components.

Answer 2

In the context of the mathematics of differential geometry, the concept of differential forms is made more precise. It is an invariant object, and does not transform under coordinate changes. In particular, they are not objects that 'inherently' come with indices, so a 2-form like the electromagnetic field tensor would be expressed by mathematicians as simply $F$. In local coordinates, one can write this as $F=F_{\mu\nu}\mathrm{d} x^\mu\wedge\mathrm{d} x^\nu$, but this is only locally true (we usually don't give a rat's ass about this in physics though, since we typically deal with only a single chart at a time in e.g. GR).

In physics, we tend to totally forget about the invariant nature of forms (or tensors more generally) and focus only on the components, e.g. $F_{\mu\nu}$ instead of $F$, which transform under coordinate transformations. This is all just fine (and is actually the historical way of doing things) but it also causes a lot of physicists to be unaware of what these forms really are, which is not so great. In conclusion: A differential form is an object that is invariant under coordinate transformations and is defined independent of local coordinates. In particular, it does not naturally have indices and writing $F$ for a 2-form is actually technically more correct than $F_{\mu\nu}$ which just denotes its components in an (arbitrary) coordinate chart. However, note that $F_{\mu\nu}\neq F$, so we can never use $F$ as just a shorthand.

Answer 3

As Danu said, indices are not "natural" part of tensor fields, they are just a pretty outdated formalism of dealing with them. Unfortunately or fortunately, it is also a pretty well-working and efficient formalism, at least in general relativity. Electrodynamics and classical mechanics would be better off using differential form notation imo...

Anyways, the expression $$ F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu $$ has no immediate generalization to index-free formalism that also preserves this equation's general form. The $\partial$ operator involves differentiation of components, and the result is generally nontensorial, (even if partials are often used in coordinate component-based treatment) thus it has no index-free form.

The above expression is the coordinate component form of a differential operator defined on differential forms called the exterior derivative, so the above equation looks like $$ F=\mathrm{d}A $$in index-free notation, where $\mathrm{d}$ is the exterior derivative.