9
$\begingroup$

The textbook describes pseudovector like this:

Let $a,b$ be vectors and $c=a\times b$, $P$ be the parity operator. Then $P(a)=-a,P(b)=-b$ by definition. But $P(c)=c$ since both $a$ and $b$ reverse direction. So $c$ is a pseudovector.

I can't understand this. For I know from my math course that $c$ is perfectly a regular vector. So $P(c)$ should be equal to $-c$ by definition. In the example above, however, the text seems to have assumed $P(c)=P(a)\times P(b)$, which is nonsense if $c$ is a self-governed entity. What I can see is that $c$ depends on $a$ and $b$ and any transformation on $c$ must be computed after $a$ and $b$ have changed accordingly. Thus I tend to define $c:=<\times,(a,b)>$ and $P(c)$ should actually be written as $\tilde{P}(c)$ where $\tilde{P}(<f,(a,b)>)=P(f(a),f(b))$. As a result, I think a pseudovector is actually a function of vectors equipped with preallocated arguments and every transformation on a vector has a counterpart on a pseudovector like $\tilde{P}$.

Am I right?

$\endgroup$
4
$\begingroup$

You can also see it in the context of group representation theory.

$O(n)$ is generated by the same infinitesimal generators as $SO(n)$ plus the discrete generator $\mathbf P$ which is parity. The parity operator generates a discrete subgroup $\mathbb Z/2\mathbb Z$ in $O(n)$).

So if we choose a representation $\rho_{SO}$ of $SO(n)$ and a representation $\rho_P$ of $\mathbf P$, we can construct a representation $\rho$ of $O(n)$:

  • If the parity operator is represented trivially $\rho(\mathbf P)=1$, we get a "pseudo representation" (pseudoscalar, pseudovector, pseudotensor etc.).
  • If however $\rho(\mathbf P)=−1$, we get a "normal" rep (scalar, vector, tensor etc).

If you answer the question What is a spinor? with A spinor is an object that transforms in the fundamental representation of $SU(2)$, then the answer to your question is

A pseudoscalar/pseudovector/pseudotensor is an object that transforms in the $O(n)$ representation which represents a) the parity operator trivially, and b) the $SO(n)\subset O(n)$ subgroup in a scalar/vector/tensor representation.

Note that this works only for odd $n$.

$\endgroup$
4
$\begingroup$

This all links back to the fact that any time you speak about a transformation, you require all the relevant physics to transform along.

E.g. velocity would not transform the way it transforms without requiring the trajectory $x^i(t)$ it is tangent to to transform also. In the sense of the notation you use we have $v \equiv \left\langle x^i(t),\frac{\mathrm d}{\mathrm dt}\right\rangle$ and parity and other transformations would then act on this pair. Hence, when we say something "transforms" a certain way, it is actually a statement about the relationship to the objects it is derived from and their own transformations$^\dagger$. The stated applies also to your example with the cross product.

By the way, a pseudovector can be considered an entity on it's own without any need to refer to the cross product. A good example is the angular velocity vector. But in that very article you can see it is actually quite practical to use an antisymmetric angular velocity matrix. The equivalence of the representation through an antisymmetric matrix (or "tensor") and a pseudovector is described in full abstraction by ACuriousMind in the answer already linked in the comments.


$^\dagger$ This is a simplified picture, physics is actually mostly relational, so it might not be obvious what is the "derived" and what the "original". In this sense, the "original" objects would be points in space.

$\endgroup$
3
$\begingroup$

Your conclusion that

I think a pseudovector is actually a function of vectors equipped with preallocated arguments and every transformation on a vector has a counterpart on a pseudovector like $\tilde P $.

is, I think, essentially correct; the reason parity is confusing is that we tend to drop the tilde.

More precisely, whenever parity considerations are on the table, for all practical purposes vectors and pseudovectors live on different vector spaces, which I'll call $V$ and $W$. One key fact that makes this work is the fact that the addition of a vector and a pseudovector is ill-defined, and such a combination is never used or physically relevant.

In that sense, the cross product is a bilinear function $$\times:V\times V\to W$$ whose domain is a different vector space. (Similarly, you can define the analogous operations $\times:V\times W\to V$ and $\times:W\times W\to W$, which are never in danger of confusion with each other.) The parity operation then splits into two transformations, $P_V$ on $V$ and $P_W$ on $W$, which are related to the cross product as $$ P_V(v_1)\times P_V(v_2)=P_W(v_1\times v_2). $$

However, because you never apply $P_V$ to objects in $W$ and vice versa, it is acceptable to drop the subscripts.

Now, I've been deliberately vague about what $W$ actually is. It is certainly isomorphic to $V$, and shares a good bit of structure with it (for example, you can extend the dot product to combinations from both, though that does get you a pseudoscalar). However, the versions that are useful for generalizations to higher dimensions and more complex geometries (e.g. curved spacetime) look rather different; they are nicely outlined in this answer by ACuriousMind.

In the end, though, I think that questions of the form "what is, mathematically, the object $X$?" are not really that useful; instead, the fruitful questions that really get you forward on the maths are of the form "how does $X$ behave mathematically?". The answer to that is usually a set of axioms which are enough to tipify the object's behaviour, and even uniquely specify it up to a canonical isomorphism. In that case, you can come up with constructions which prove existence, but in a sense all you need are the axioms.

One case where that happens is with tensors, where the universal property is such an axiom. (For examples, see this thread.) As far as pseudovectors go, the facts laid out above are pretty close to a complete set of axioms that constrain their behaviour closely enough to be useful.

$\endgroup$

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.