-1
$\begingroup$

Physicists often talk about a vector representation. For example, the first comment to this answer says:

That the $(1/2,1/2)$ representation corresponds to a vector irreducible representation of the Lorentz group is not obvious.

From a mathematical point of view all representation are valued in vector spaces so it's somewhat confusing to talk of vector representations. That is all reps are

$G \rightarrow Aut V$

Where $V$ is some vector space.

My first inclination is that it's simply a physicists way of speaking to distinguish between spinorial reps (which, rather than going from G, goes from the universal cover of G, which in the case of the Lorentz group is its double cover) and ordinary 'vector' representations, which goes directly from G.

Is this on the right track, or is there more to calling a rep, a vector rep?

$\endgroup$
  • $\begingroup$ Note that a vector means many different things in mathematics & physics depending on context. Related: physics.stackexchange.com/q/155878/2451 $\endgroup$ – Qmechanic Dec 3 '17 at 9:29
  • 2
    $\begingroup$ You might be overthinking this a bit... When saying $5\;\mathrm m$ and $(3\;\mathrm m\;,4\;\mathrm m)$ it is fairly convenient, although not a hundred percent semantically fitting, to call the latter a vector representation rather than the former. A quick and easily understood use of words. Or am I misunderstanding your question? $\endgroup$ – Steeven Dec 3 '17 at 9:33
  • $\begingroup$ @qmechanic: maybe so in physics; but it has a precise definition in mathematics. $\endgroup$ – Mozibur Ullah Dec 3 '17 at 9:41
  • 2
    $\begingroup$ Because for "non-mathematicians" like me, the term "vector" is intuitively understood as an arrow. 5m is therefore different from any multiple coordinate representations such as (3m,4m), because they can be drawn (or at least imagined) as arrows on a piece of paper. $\endgroup$ – Steeven Dec 3 '17 at 10:26
  • 1
    $\begingroup$ @steeven: mathematically speaking they're in the dual vector space, so they're vectors; but again, I see what you're driving at - they're not 'physical' vectors. $\endgroup$ – Mozibur Ullah Dec 4 '17 at 6:28
2
$\begingroup$

A similar thing happens in mathematics as well. Although $V\otimes \cdots \otimes V \otimes V^* \otimes \cdots \otimes V^*$ is a vector space, in many contexts its elements are not called vectors but tensors, whereas the name vector is reserved for elements of $V$.

For the Lorentz group we have, for example, the trivial representation, which we call a scalar (even if it is a vector, trivially). The spinor representations are best identified as spinors rather than as vectors (although they are vectors, of course). We can use the name vector for the representation $(1/2,1/2)$, because it acts on the vectors in the physical space-time. For higher spin, we have other specific names to identify the representations. For example, just as in mathematics, the tensor products of our vector representation are usually called tensors.

$\endgroup$
  • $\begingroup$ +1: this suggests that spinors don't act on space time, is that right? $\endgroup$ – Mozibur Ullah Dec 3 '17 at 12:25
  • 1
    $\begingroup$ @MoziburUllah Yes, or being a bit more precise: the spinor representations do not correspond to the space-time or it tangent spaces $\endgroup$ – coconut Dec 3 '17 at 12:31
  • $\begingroup$ It is misleading to even wrong to think/say that spinors (better said spinorial fields) do not act on flat/curved spacetime. The concept of spinor bundle (briefly touched upon by R. Wald in his chapter 13) explains everything, starting with the SL(2,C) principle fiber bundle over spacetime. $\endgroup$ – DanielC Dec 3 '17 at 21:23
  • $\begingroup$ @DanielC Hmm, it seems strange to me to even talk about spinors (or spinorial fields) acting on anything. What do you mean, exactly? $\endgroup$ – coconut Dec 3 '17 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.