Are $v^ie_{i}$ and $v^iv_{i}$ (where $v$ are the components and $e$ the basis vectors) both tensors? Or only the second one?

Question

I am studying the math of tensors, I have an understanding of the concepts of covariance, contravariance, dual spaces, Einstein notation and so on. I am a bit confused about the notation though. My quick questions:

• Is $v^ie_{i}$ a tensor? ($v$ are the components of a vector and $e$ are the basis vectors)

• Is $v^iv_{i}$ a tensor?

• If both the above are tensors, how do you distinguish between the two? If not, isn't this an abuse of notation?

• When talking about a tensor, does one usually mean $v^iv_{i}$, $v^ie_{i}$, or both? (I guess this depends on the answer to the first two points)

Answer 1

Scalars are tensors of rank 0.

Vectors are tensors of rank 1.

Both the objects you describe are therefore tensors. The main difference is notation.

$v^i e_i$ might be better written $v^i \mathbf e_i$. This is a vector, in which the basis vector $\mathbf e_i$ is explicit. In index notation, we omit to write the basis vectors. The same vector is written $v^i$ in index notation. For almost all practical purposes index notation is clearer and makes calculation easier. We can simply write $v^i$, meaning the vector $\mathbf v = v^i \mathbf e_i$

$v^i v_i$ is already written in index notation. It is a scalar, equal to $\mathbf v \cdot \mathbf v$ (note: there is no abuse of notation here, the dot product is indeed calculated by summing the products of components).

More generally, we deal with higher rank tensors. E.g. $v^i v_j$ is a rank $2$ tensor written in index notation (again there is no abuse of notation, because a tensor is specified by its components in a given basis).

Answer 2

There is a lot of abuse of notation in GR I find. I would add something specifically about scalars:

$v^i v_i$ is a scalar because it is shorthand for an inner product between two elements of the vector space. An inner product in this scenario is a map which takes two vectors and gives you a number (in $\mathbb{R}$ or $\mathbb{C}$ usually). Let's call the map $\langle,\rangle$:

$\langle,\rangle : V \times V \rightarrow \mathbb{C}$

$\langle,\rangle : (\mathbf{v},\mathbf{w}) \mapsto \langle \mathbf{v},\mathbf{w} \rangle = v^i w_i$

where Einstein summation has been used to indicate that the way the map works is to elementwise-multiply the components of the two vectors to yield a simple number (a scalar). The placement of the indices indicates another way of thinking about the inner product: since the dual vector space is in fact defined as the space of maps from the vector space to $\mathbb{C}$, what an inner product can be thought of as doing is taking one of the vectors (say $\mathbf{v}$), tracking down the dual vector with the same coordinates in the dual space ($\tilde{\mathbf{v}}$), and using that as a map on the other vector (which one it finds the dual of doesn't matter since the inner product is symmetric):

$\langle \mathbf{v},\mathbf{w} \rangle = \tilde{\mathbf{v}} (\mathbf{w}) = \tilde{\mathbf{w}} (\mathbf{v}) $

Coordinates of course depend on the basis you're in, but in GR (at least at first) we are often interested in basis transformations which leave the inner product invariant. This is why 'scalar' is often used to mean 'an object which is not changing under the basis transformations I am doing'.

I myself would indeed refer to $\mathbf{v}$ as a vector, since it sits in the vector space (actually the tangent space in GR), expressible as $v^i \mathbf{e}_i$ with the knowledge that its components change "as a vector" (as they say). The reason that the components of covectors change in the opposite way under basis transformations is by design, so that the inner product does not change (the changes cancel each other out). A vector is a type of tensor. To form higher-dimensional tensors you can take a type of product called a tensor product between lower-dimensional tensors (I won't address that here).

$v_i \mathbf{\theta}^i$ is the label given to the dual vector in the dual space whose numerical coefficient values with respect to a special basis are the same as the vector $v^i \mathbf{e}_i$. This dual basis $\{\theta^i\}$ is defined to be the one dual to whichever one you have chosen in your original vector space i.e. the $i$ maps that satisfy $\theta^i (\mathbf{e}_i) = \delta_i^j$.

I've definitely abused a bunch of notation in this answer, but I hope it yielded some information for you! People use 'tensor' and 'vector' to refer both to the original objects in their respective spaces (coordinate-independent objects) or to the numerical coordinates with respect to some basis. It depends on level of mathematical rigour needed, plus personal taste, I believe.

Answer 3

• $v^ie_i$ is an element of the vector space, but it's not a vector in the same sense as $v_i$. This is because we define vectors/tensors based on their transformation properties.
• $v_iv^i$ is a scalar.
• Points addressed above
• When talking about tensors, we usually just mean the components $v_i$ (In this case, the $v_i$s transform as a vector)

We usually have a vector space with basis $e_i$ and each element of the vector space $A$ can be expanded as $\vec{V} = v^ie_i$. The $v^i$ completely determines $\vec{V}$ in a given basis, and it is these components that transform as a vector. Note that under a change of basis $\vec{V}$ itself remain unchanged and that's how we get the rules of transformation for $v^i$ from that of $e_i$.

We get dual space by realizing that the space of all linear functionals $T:A\to \mathbb{R}$ themselves form a linear vector space. As a result we can expand each element of $T$ along a basis $e^j$ defined by $e^je_i = \delta_{ij}$. This is the dual space.

We get higher order tensors by taking the tensor product $A^n \otimes T^m$ and each element $r$ can be expanded along the basis of this space. The coefficients are what transforms as a tensor and that's what we're interested in.

Sean Carroll's GR book has a nice section on the whole construction of the dual space and tensor product spaces in a crisp manner.