# A Abstract tensor notation

1. May 5, 2016

### JonnyG

In Wald's "General Relativity", in his section on abstract tensor notation, he lets $g_{ab}$ denote the metric tensor. When applied to a vector $v^a$, we get a dual vector, because $g_{ab}(v^a, \cdot)$ is just a dual vector. Okay cool. But then he says that this dual vector is actually $g_{ab}v^b$, which is a contraction. But don't we have $g_{ab}v^b = \sum\limits_{i=1}^n g(\cdot, v^i) = \sum\limits_{i=1}^n g_{ab}(v^i, \cdot)$, which in general is not going to be equal to $g_{ab}(v^a, \cdot)$? Where am I messing up here?

EDIT: $\{v^i\}$ is a basis for the tangent space.

Last edited: May 5, 2016
2. May 5, 2016

### Fightfish

Well, the metric tensor is a symmetric tensor, so the order of the arguments doesn't matter.

3. May 5, 2016

### JonnyG

I know, but suppose for example $v^a = \sum\limits_{i=1}^n c_i v^i$ for scalars $c_i > 1$. Then $g_{ab}(v^a, \cdot) \neq \sum\limits_{i=1}^n g_{ab}(v^i, \cdot)$

EDIT: My issue is, when you apply the metric tensor to the vector $v^a$ then you get the dual vector $g_{ab}(v^a, \cdot)$. But Wald writes that this is also the tensor $g_{ab}v^b$. Now, I don't see how $g_{ab}(v^a, \cdot) = g_{ab}v^b$. In fact, according to my calculations above, they are not, in general, equal.

EDIT 2: Hold on, is $g_{ab}v^b$ really a contraction? I thought contractions were applied to a single tensor? I mean, you can contract $T^{abc}_{de}$, but I don't see how $g_{ab}v^b$ is a contraction.

Last edited: May 5, 2016
4. May 5, 2016

### Fightfish

Ah okay I get what you are trying to say now; the notation used in that book is really really terrible. Let me switch to friendlier notation: using an overbar for vectors and a tilde for dual vectors (also known as one-forms).
So, we have
$$\tilde{V} = \mathbf{g} (\bar{V}, \cdot)$$
To find out the components of this dual vector, we evaluate
$$V_{i} = \tilde{V}(\bar{e}_{i}) = \mathbf{g} (\bar{V}, \bar{e}_{i}) = \mathbf{g} (V^{j} \bar{e}_{j}, \bar{e}_{i}) = V^{j} g_{ij}$$

As for the strange notation of $g_{ab}(v^{a}, \cdot)$, I have absolutely no idea what it means...when you put superscripts or subscripts you are referring to the components already evaluated in some basis.

5. May 5, 2016

### Fightfish

If $v^a$ is a general vector, why does it have a superscript - and why does it share the same index as that of the metric tensor? I think the source of your confusion stems from the notation.
Yes, in standard notation, it is a contraction. You can always define $T_{ab}^{c} \equiv g_{ab}v^c$, no?

6. May 5, 2016

### JonnyG

I will type exactly what the section says in my textbook. It's only a few lines:

Additional notational rules apply to the metric tensor, both in the index and component notations. Since a metric $g$ is a tensor of type $(0,2)$, it is denoted $g_{ab}$. If we apply the metric to a vector, $v^a$, we get the dual vector $g_{ab}v^b$.

7. May 5, 2016

### JonnyG

I would like to go over some things in regards to this abstract tensor notation..Suppose we have a tensor of type $(3,2)$. We can denote it by $T^{abc}_{de}$. This means that $T: V^* \times V^* \times V^* \times V \times V$. So, let $\{\phi^*_i\}$ denote the elementary contravariant tensors on $V$ and let $\{\phi_i\}$ denote the elementary covariant tensors on $V$. Then we can write $T^{abc}_{de} = \sum\limits_{i,j,k,u,v = 1}^n \phi^*_i \otimes \phi^*_j \otimes \phi^*_k \otimes \phi_u \otimes \phi_v$. Do we write $T$ is equal to this summation? Or do we write $T^{abc}_{de}$ is equal to this summation? Because at first I thought $T^{abc}_{de}$ denoted the tensor, but now I am starting to think that it just denotes the general component of the tensor?

EDIT: Sorry, it should be $T^{abc}_{de} = \sum\limits_{i,j,k,u,v = 1}^n \phi_u \otimes \phi_v \otimes \phi^*_i \otimes \phi^*_j \otimes \phi^*_k$ right?

8. May 5, 2016

### Fightfish

After racking my brain, I honestly have no idea what Wald is saying, although I guess that whatever he is trying to say is essentially equivalent to whatever I posted in #4. More accurately, $g_{ab}V^b = V_{a}$ is the a-th component of the dual vector $\tilde{V} = \mathbf{g}(\bar{V}, \cdot)$.

Well, it can sometimes mean the former when used loosely, i.e. the tensor $T^{abc}_{de}$ blahblahblah, when when used mathematically, it refers to the component - after all, when you perform manipulations and summations, you are working with numbers (i.e. components).

I'm not sure that this makes any sense. We usually write
$$\mathbf{T} = T^{abc}_{de} \bar{e}_{a} \otimes \bar{e}_{b} \otimes \bar{e}_{c} \otimes \tilde{\omega}^{d} \otimes \tilde{\omega}^{e}$$
where $\bar{e}$ and $\tilde{\omega}$ are the basis vectors and dual vectors (one-forms).

9. May 5, 2016

### George Jones

Staff Emeritus
Consider tensor products of the vector space (over the reals) $V$ and its dual space

In Penrose's abstract index notation, which Wald uses, and which I dislike, an index that uses the latin alphabet labels the space in which the object lies, so $v^a$ is an element of $V$, not a real number (component with respect to a basis), $g_{ab}$ is an element of $V^* \otimes V^*$, etc. Latin indices unambiguously give the space in which the object lives.

Indices that uses the greek alphabet specify components with respect to a particular basis, i.e., are real numbers.

$g_{ab} v^b$ lives in $V^*$ (not in $\mathbb{R}$), and in more traditional notation might be written as a contraction of $\bf{g}\otimes\bf{v}$