# Show that the contraction of a covector and a vector is Lorentz invariant

I just got Sean Carroll's Spacetime and Geometry: An Introduction to General Relativity a couple of weeks ago, and I have resolved to go through the entire book. In the first chapter, he prompts the reader to check that $$\omega_\mu V^\mu$$ is Lorentz invariant by using the transformation laws of vectors and their dual covectors.

When I read this, I assume he wants the reader to show that $${\Lambda^{\upsilon'}}_\mu \omega_\mu V^\mu = \omega_{\upsilon'} V^{\upsilon'}$$ after a Lorentz transformation has been applied to the left side. However I am confused when applying a Lorentz transformation to the LHS, because in the book Carroll shows that covectors transform according to the inverse of the transformation used on vectors, so I have 2 questions:

1. How do I obtain $$\omega_{\upsilon'}$$ from $${\Lambda^{\upsilon'}}_\mu \omega_\mu$$?

2. If question 1 is meaningless or not well posed, my main question is how do I show that the action of a covector on a vector is Lorentz invariant, using only the transformation properties of vectors and covectors?

$$s=\omega_\mu V^\mu$$ is scalar. It has "no" indices; the $$\mu$$ that looks like an index is internal to the definition of $$s$$ (it is summed over), in the same way that $$c=\int_0^5x\,dx$$ is a constant with no dependence on the variable $$x.$$

There are two ways to transform $$\omega_\mu V^\mu$$ by a Lorentz transform $$\Lambda.$$ One way is to transform the quantity as a whole. The rule is, for every free index $$\mu$$ in the quantity, to add a factor of $${\Lambda^{\mu'}}_\mu$$ (or $${\Lambda_{\mu'}}^\mu$$). But $$\omega_\mu V^\mu$$ has no free indices, so it doesn't change at all: $$\omega_\mu V^\mu$$ transformed by $$\Lambda$$ is $$\omega_\mu V^\mu.$$ I.e. "scalars are invariant."

The other way is to transform each of its parts. $$V^\mu$$ has one up index, so transforms (by the above generic rule!) to $${\Lambda^{\mu'}}_\mu V^\mu.$$ $$\omega_\mu$$ has one down index so transforms to $${\Lambda_{\mu'}}^\mu\omega_\mu.$$ I will rename $$\mu$$ to $$\nu$$ in the previous sum to avoid clashing, and then substitute both transformations into $$s$$ to find that $$\omega_\mu V^\mu$$ transformed by $$\Lambda$$ is also $${\Lambda^{\mu'}}_\mu{\Lambda_{\mu'}}^\nu\omega_\nu V^\mu.$$

Your task is to show these two versions of the transformation to be consistent (that applying the vector and covector transformation laws agrees with the scalar (lack of) transformation law): $${\Lambda^{\mu'}}_\mu{\Lambda_{\mu'}}^\nu\omega_\nu V^\mu=\omega_\mu V^\mu$$

• So the "rule" just adds the convenient form of the Lorentz transform to whatever it is transforming? I thought the transform was some kind of function that did something to vectors and covectors. Assuming this rule, is ${\Lambda^{\upsilon'}}_\mu \omega_\mu$ a meaningless operation? Apr 30, 2022 at 5:38
• The transform is a function that "does something" to by adding the "convenient form" of the Lorentz transform. They not mutually exclusive concepts. If you would like it explicitly, you could consider $f^\mu(V)={\Lambda^\mu}_\nu V^\nu$ and $f_\mu(\omega)={\Lambda_\mu}^\nu\omega_\nu$ to define the transformation of $V$ to $f(V)$ and $\omega$ to $f(\omega).$ I don't think anyone bothers with this notation. Yes, ${\Lambda^{\nu'}}_\mu\omega_\mu$ is meaningless, since $\mu$ appears twice, both down. An index in a term must appear either as a pair (one up, one down) or just once.
– HTNW
Apr 30, 2022 at 5:49
• Alright, that cleared it up for me, thank you. Apr 30, 2022 at 5:55