# Order of index in Lorentz transform

I am reading Schwartz's "QFT and the standard model". On pg 13 he gives the Lorentz transform of a rotation around the x-axis:

$\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos \theta _x & \sin \theta _x \\ 0 & 0 & -\sin \theta _x & \cos \theta _x \\ \end{array} \right)$

For a boost along the x-axis he gives:

$\left( \begin{array}{cccc} \cosh \beta _x & \sinh \beta _x & 0 & 0 \\ \sinh \beta _x & \cosh \beta _x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$

I believe this corresponds to the Lorentz transform in the form $\Lambda^\alpha{}_\beta$. I believe the first index is the row and the second is the column. But at the bottom of the pg he has $V^\mu=\Lambda_\nu^\mu V^\nu.$ as though the order of the indices on $\Lambda$ doesn't matter. But surely they do matter as $\Lambda^\alpha{}_\beta\not=\Lambda_\beta{}^\alpha$ in general?

• This question seems to be about a typographic detail in a specific book. – zzz Jun 6 '15 at 2:51
• @bechira I think it would be interesting to know as its otherwise such a good qft textbook. I think the answer is by $\Lambda^\mu_\nu$ he means $\Lambda^\mu{}_\nu$. – Virgo Jun 6 '15 at 3:59

His notation is not ambiguous because the expression $$V^{'\mu} \equiv \Lambda^\mu_\nu V^\nu$$ can only mean sum along the $\nu$ component. Since $\Lambda$ is a representation of the Lorentz group, it is a linear operator, hence it can only act on a vector by the usual way that matrices act on vectors. Hence the above is unambiguous.

I'll explain why in general this notation is unambiguous.

The convention is the following: the upper index denotes a vector index and the lower index denotes a dual vector index and. That is, for a general rank $(p, q)$ tensor, one would usually write

$$T^{\mu_1...\mu_p}_{\nu_1...\nu_q} \partial_{\mu_1} \otimes ... \otimes \partial_{\mu_p} \otimes dx^{\nu_1}... dx^{\nu_q}$$

where $\{\partial_\mu\}$ denotes a basis for your vector space and ${dx^\nu}$ denotes a basis for its dual vector space. (The notation I chose is from differential geometry, which is certainly useful when you define a field theory in over a general spacetime manifold).

As a special case, a linear transformation is a rank $(1,1)$ tensor. Namely $\Lambda^\mu_\nu$ unambiguously denotes components of the tensor:

$$\Lambda^\mu_\nu ~\partial_\mu \otimes dx^\nu$$

Acting on a vector $V \equiv V^\tau \partial_\tau$, we get:

$$\Lambda[V] \\ = \Lambda^\mu_\nu ~\partial_\mu \otimes dx^\nu [V^\tau \partial_\tau] \\ = \Lambda^\mu_\nu V^\tau (\partial_\tau dx^\nu) ~\partial_\mu \\ = \Lambda^\mu_\nu V^\tau \delta_\tau^\nu ~\partial_\mu \\ = \Lambda^\mu_\nu V^\nu ~\partial_\mu$$

Observe no horizontal padding in the indices are needed to make these manipulations unambiguous.

• but $\Lambda^\mu_{\ \nu}$ is not a tensor, right? – Nahc Jan 21 '17 at 16:39
• @Chan Why not? It's a well-defined multi linear operator in any basis. – zzz Jan 22 '17 at 19:51