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I am trying to understand the meaning of upper and lower indices as used in the Newman-Penrose formalism.

The tetrad is $\lbrace l^{a},n^{a},m^{a},\overline{m}^{a}\rbrace$, where the upper index means that its a vector (i.e. column vector) in $\mathbb{R}^{4}$.

Consequently, $l_{a}$ should be the transpose of $l^{a}$, right? That is, $(l^{a})^{T}=l_{a}$ (i.e. row vector).

Now, for instance, the product $l^{a}n_{a}$ is defined since it's like a scalar product, right? We can multiply column and row vector.

However, I came across the following formula for a bivector:

$$U_{ab}=-l_{a}m_{b}+l_{b}m_{a}$$

I am utterly confused by this formula. How do I understand the right hand side? I have a multiplication of two row vectors, how is it even defined? Further, the indices are not the same in this "row vector product", so I cannot use the Einstein summation convention.

Can anybody explain what the right hand side explicitly means?

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First, $l_a$ is not the transpose of $l^a$. There would typically be a sign change on some components to account for the metric. This is referred to as the adjoint instead, and the transpose is equal to the adjoint only for the Euclidean metric. Compare with quantum mechanics, where you often deal with the conjugate transpose. This is also called the adjoint.

You're confusing yourself because you're thinking that $l_a$ and $m_b$ are row vectors and that the multiplication operation here is matrix multiplication. That gives you the right answer in the scalar product case, but it doesn't work here.

Consider: given a vector $v^b$, the quantity $U_{ab} v^b$ would describe a matrix $U$ acting on a vector $v$ to produce a covector. Hm, no, that doesn't work either: matrices don't ever do that.

Consider again: define ${U^a}_b = g^{ac} U_{cb}$. Now we can do ${U^a}_b v^b = g^{ac} (-l_c m_b + +l_b m_c) v^b$.

Something you should get used to: index notation doesn't rely on terms in a product being adjacent. You should infer the corresponding matrix products from indices being summed over. That's one way you could realize that $l_a m_b$ doesn't correspond to row-vector column-vector multiplication: there is no common index to sum over.

Now, how can you read this? Consider $g^{ac}(-l_c m_b) v^b$. The $g^{ac} l_c$ part turns a covector into a regular vector, so we get $l^a m_b v^b$. There's a name for this: the tensor product. It's sometimes denoted by multiplying a column vector onto a row vector. Usually, it'd be written $(l \otimes m)(v)$. Either way, you should recognize $l^a m_b$ as describing a matrix formed a column vector and a row vector. Again, in index notation, order is not important. $m_b l^a$ corresponds to the same matrix. It is the placement of the indices and what ultimately sums over them that matters.

In truth, I think you'll have to let go of trying to maintain a correspondence between index notation and matrix notation. Matrices will not help you when you start dealing with 3-index tensors or higher. It is helpful to keep thinking of tensors as linear transformations--it's just that, unlike matrices, these transformations can have more than one argument.

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  • $\begingroup$ thank you, it's much better now. However, there is still a part I don't quite understand: according to what you wrote, the expression $-l_{c}m_{b}+l_{b}m_{c}$ only makes sense when multiplied in the front by a metric. Then we get a regular vector and in the end the tensor product. However, in my definition there was no metric anywhere. In particular, how could I make sense out of smth. like $U_{ab}U_{cd}$? @Muphrid $\endgroup$
    – user46446
    Commented Dec 3, 2015 at 8:23
  • $\begingroup$ Well, therein lies the problem with trying to think about them as matrices: you're not equipped to deal with something that converts a vector to a covector, let alone two such somethings. - But a general concept of tensors as functions of vectors and covectors (functions that are linear in their arguments) makes it easy: you have a function of four vectors, which is some other function taking the first two vectors as arguments, multiplied by that same function taking the second two vectors as arguments. $\endgroup$
    – Muphrid
    Commented Dec 3, 2015 at 8:27
  • $\begingroup$ I see, the argument with the functions makes sense. But assume I have a tetrad given explicitly, i.e. I have four explicitly given vectors $l^{a},n^{a},m^{a}$ and $\overline{m}^{a}$ (assume it's a null tetrad, for simplicity). Given these expressions, how do I calculate $U_{ab}$? $\endgroup$
    – user46446
    Commented Dec 3, 2015 at 8:35
  • $\begingroup$ Common practice is to compute it one component at a time. From the function standpoint (the standpoint of tensors as "multilinear maps"), $U_{ab}$ means a function of two vector arguments, so you plug in basis vector for each argument and evaluate. You would need the tetrad vectors' components with respect to that basis. You could also use the tetrad itself as a basis--the inner products of the tetrad vectors (at least, those used in Newman-Penrose) are known values by construction. $\endgroup$
    – Muphrid
    Commented Dec 3, 2015 at 8:41
  • $\begingroup$ Ah ok, so is it correct if think like this: $l_{a}$ is the $a$th component of the vector $g_{ab}l^{b}$? $\endgroup$
    – user46446
    Commented Dec 3, 2015 at 8:48
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indices can be raised and lowered by the global metric $g_{ab}$ which can be obtained via $$g_{ab}=-l_{a}n_{b}-n_{a}l_{b}+m_{a}\bar m_{b}+\bar m_{a}m_{b}$$ with $g^{ab}=(1/g_{ab})$

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