Confused about basis vector notation

[snoopies622]

Why are basis vectors represented with subscripts instead of superscripts? Aren’t they vectors too? Isn’t a vector a linear combination of basis vectors (and not basis co-vectors?)

In David McMahon’s Relativity Demystified, he says,

“We will often label basis vectors with the notation [itex]e_a[/itex]. Using the Einstein summation convention, a vector V can be written in terms of some basis as [itex] V=V^{a}e_{a}[/itex]. In this context the notation [itex]e_a[/itex] makes sense, because we can use it in the summation convention (this would not be possible with the cumbersome [itex] (\hat{i}, \hat{j}, \hat{k} ) [/itex] for example).”

But using the Einstein summation convention, [itex] V=V^{a}e_{a}[/itex] is the inner product of a vector and a co-vector, which is a scalar and not a vector at all.

 

[axmls]

But using the Einstein summation convention, V=Vaea V=V^{a}e_{a} is the inner product of a vector and a co-vector, which is a scalar and not a vector at all.

I'd like to see what someone more knowledgeable says, but I'd like to point out that the correct notation would be ##V = V^a \vec{e}_a##, not ##V = V^a e_a##, i.e. upper indices are used for components of vectors, but the indices on the basis are labeling four distinct vectors, not vector components. In fact, to represent a particular component of a basis vector, I believe I've seen the notation ##(\vec{e}_a)^b##.

Of course, there might be a more convincing reason, and I'd like to see others' responses to this, but obviously the lower index allows the summation convention to operate, so that would be one reason.

 

[snoopies622]

hmm . . So in that case, [itex] \vec{e_a} [/itex] would actually be a (1,1) tensor. Surprising, but it makes sense. I've seen just [itex] e_a [/itex] for basis vectors elsewhere too, but this is enough for me to believe that in every case that representation is incorrect. Thanks, axmls.

 

[zinq]

One early and extremely widespread version of the Einstein summation convention is to sum over repeated indices, but only if one is raised and the other is lowered (i.e., one is a superscript and the other is a subscript).

In the expression asked about:

V = V^α e_α,

the e_α are vectors, so the expression would not even make sense as an inner product (or equivalently, as a covector operating on a vector).

 

[stevendaryl]

hmm . . So in that case, [itex] \vec{e_a} [/itex] would actually be a (1,1) tensor. Surprising, but it makes sense. I've seen just [itex] e_a [/itex] for basis vectors elsewhere too, but this is enough for me to believe that in every case that representation is incorrect. Thanks, axmls.

Hmm. I don't see how it makes sense to view basis vectors as a [itex](1,1)[/itex] tensor. [itex]e_a[/itex] is just a labeled vector, and [itex]V^a e_a[/itex] just means [itex]V^0 e_0 + V^1 e_1 + V^2 e_2 + V^3 e_3[/itex]. It's definitely not any kind of product of a vector and a tensor. The result is not a scalar, it's a vector.

Here's the more general rule: If you have a legal expression with each of the upper-indices matched by a corresponding lower index (and summed over), the result is a geometric object that is independent of your choice of basis. If you look at the expression

[itex]V^a e_a[/itex]

It has two basis-dependent parts: [itex]V^a[/itex] will change depending on your basis, and so will [itex]e_a[/itex], but the combination [itex]V^a e_a[/itex] (summed over [itex]a[/itex]) has the same value in every basis (but the value is a vector, not a scalar).

This is a different point of view than the usual one that people are taught about vectors. People are often told something like "a 4-vector is a set of 4 numbers [itex]V^0, V^1, V^2, V^3[/itex] which transform in such-and-such a way under a change of coordinates". To me, it's more helpful to think that a 4-vector [itex]V[/itex] is a geometric object that is independent of coordinate systems, and the only role of a coordinate system if for picking a way to write that vector as a linear combination of other vectors: [itex]V = V^0 e_0 + V^1 e_1 + V^2 e_2 + V^3 e_3[/itex]. [itex]V[/itex] is not just the 4 numbers [itex]V^a[/itex], but is the linear combination [itex]V^a e_a[/itex].