# Group representations and active/passive transformations

Suppose we are in Euclidean 3-space with coordinates $x$ and a scalar function $\phi(x)$ defined on it, and consider the group of rotations $SO(3)$ for simplicity. Take a rotation matrix $R \in SO(3)$; then the usual explanation for the difference between active and passive transformation is:

• Active: If we picture the function $\phi(x)$ as having a "bump" somewhere, an active transformation moves this bump around. We now have a new function $\phi'$ of the same coordinates $x$, given by $\phi'(x) = \phi(R^{-1}x)$.

• Passive: We leave the bump fixed and instead make $R$ rotate the coordinate axes to get a new set of coordinates $x'$ on the same physical space. In terms of the new coordinates, the field is expressed as a function $\phi'(x') = \phi(Rx')$ or $\phi'(x) = \phi(Rx)$ if you prefer, since the name of the coordinates doesn't make any difference.

This makes perfect sense but there's a problem: the passive transformation is not a representation of the group of rotations. If we are to think of transformations as an action of $SO(3)$ (or the Lorentz group or whatever your favorite group is) on the space of functions, we can only use active transformations, since only they are actually a group action. Does this imply that if we study fields as representations of some group we are restricting ourselves to active transformations only? Shouldn't the active and passive viewpoints be equivalent?

Edit: the passive transformation is not a group representation because if we define $(\rho_R \phi)(x) = \phi(Rx)$, we get $\rho_{R_1} \rho_{R_2} = \rho_{R_2 R_1}$ instead of the other way around.

• Great question. This is something I realized years ago, it bothered me for a few hours, then I got distracted and forgot about it, and now I feel perturbed again thanks to you (but it's a great kind of perturbed of course). In practice, I'm not sure I've seen passive transformations used much or at all in field theory which perhaps explains why one can get away with not thinking deeply about this. – joshphysics Apr 23 '18 at 16:45
• Is a passive transformation any different from a gauge transformation? – Ryan Thorngren May 8 '18 at 17:39
• @RyanThorngren As far as I can tell, yes. Active/passive is a perspective that is independent of the kind of transformation considered. You can see that I'm talking about global transformations in my post, not local. – Javier May 8 '18 at 17:40
• I'm confused as to why you say that the passive transformations are not representations of the group action. I would say they are such a representation, and they rotate the coordinate axes. What's wrong with this thinking? – kleingordon May 8 '18 at 17:44
• We can make a local change of coordinates and it also doesn't affect the physics, using polar coordinates, for instance... – Ryan Thorngren May 8 '18 at 17:46

## Active vs. Passive.

An active transformation is a map $T\colon \mathrm O(n)\to \mathrm{End}(\mathscr C(\mathbb R^n))$ defined as $$(T_R\phi)(x):=\phi(R^{-1}x)\tag1$$

A passive transformation is a map $f\colon\mathrm O(n)\to \mathrm{End}(\mathbb R^n)$ defined as $$f_R(x):=Rx\tag2$$

These two view points are related through $$T_R\phi\equiv\phi\circ f_R^{-1}\tag3$$

They both furnish representations of $\mathrm O(n)$: $$T_R\circ T_{R'}=T_{RR'},\qquad f_R\circ f_{R'}=f_{RR'}\tag4$$ as is trivially checked: \begin{aligned} (T_R(T_{R'}\phi))(x)&\overset{(1)}=(T_{R'}\phi)(R^{-1}x)\overset{(1)}=\phi(R'^{-1}R^{-1}x)=\phi((RR')^{-1}x)\\ f_R(f_{R'}x)&\overset{(2)}=f_R(R'x)\overset{(2)}=RR'x \end{aligned}\tag5 as required.

Needless to say, and because of equation $(3)$, establishing the group structure of $T$ automatically implies that of $f$ and vice-versa. We included both calculations to illustrate both viewpoints.

## Left vs. Right.

OP is actually asking about a non-standard notion of active vs. passive, that I will call left vs. right for lack of a better term. I will focus on $T$, but the discussion in terms of $f$ is analogous.

A left (active) transformation is a map $T\colon \mathrm O(n)\to \mathrm{End}(\mathscr C(\mathbb R^n))$ defined as $$(T_R\phi)(x):=\phi(R^{-1}x)\tag6$$

A right (active) transformation is a map $\tilde T\colon \mathrm O(n)\to \mathrm{End}(\mathscr C(\mathbb R^n))$ defined as $$(\tilde T_R\phi)(x):=\phi(Rx)\tag7$$

Unlike before, these two transformations are two different transformations, not two different points of view of the same transformation. That being said, they are related through $$T=\tilde T\circ \mathrm{inv}\tag8$$ where $\mathrm{inv}\colon R\mapsto R^{-1}$ is the inverse map.

As discussed before, $T$ is a $G$-homomorphism, where $G$ is the group $G=(\mathrm{End}(\mathscr C(\mathbb R^n)),\circ)$. In other words, $T$ is a representation of $\mathrm O(\mathbb R^n)$, with $\mathrm{End}(\mathscr C(\mathbb R^n))$ being the representation space. The group product is just composition.

On the other hand, $\tilde T$ is a $G$-anti-homomorphism (indeed, $\tilde T$ is the composition of a homomorphism $T$ and the standard (involutive) anti-homomorphism $\mathrm{inv}$, cf. $(8)$). In other words, $\tilde G$ it is not quite a representation, in accordance with OP's claims. One should point out that $\tilde T$ is a $\tilde G$-homomorphism, where $\tilde G$ is the group $\tilde G=(\mathrm{End}(\mathscr C(\mathbb R^n)),\star)$, where the group product is defined as $\star=\circ\otimes\tau$, where $\tau\colon A\otimes B\mapsto B\otimes A$ is the swap map. In other words, $\star$ is the product $$A\star B:=B\circ A\tag9$$

That $\tilde G$ is indeed a group is an easy exercise that is left to the reader. This group is typically known as the opposite group of $G$, and is denoted by $G^\mathrm{op}$.

The conclusion is now immediate: $\tilde T$ is also a representation of $\mathrm O(n)$, but the group product is not $\circ$, but $\star$ instead.

• Maybe this is just me being confused, but I would call both those transformations active, though one acts on the coordinates and the other on the field. Suppose $R$ is a counter-clockwise rotation around the z-axis. Then $T_R \phi$ is a function whose profile has been rotated in the same direction as $R$; that's an active transformation. A passive transformation would imply that $R$ rotates the coordinate axes, and we seek a new coordinate expression for $\phi$. – Javier May 8 '18 at 17:52
• To me, active and passive transformations should both act on the same space; you define one of them on $C(\mathbb{R}^n)$ and the other on $\mathbb{R}^n$ . Now that I think about it, they can both act on $\mathbb{R}^n$ as well as $C(\mathbb{R}^n)$: an active transformation rotates a vector so that its components change to $Rx$, while a passive transformation rotates the axes and leaves the vector fixed, which changes its components to $R^{-1}x$. I think the left/right action is probably the way to interpret it, like you (and the other user) say. – Javier May 8 '18 at 18:16
• @Javier I updated the answer. I hope everything is clear now. If something isn't, just leave a comment. Cheers! – AccidentalFourierTransform May 10 '18 at 13:21
• I like this answer a lot. I still think the standard meaning of active/passive is the one I gave, but let's not argue about that, it's not super important. I assume the standard theory of group representations is also valid for right actions, right? – Javier May 10 '18 at 13:29
• @Javier tomayto, tomahto, right? thank you btw :-) Yes, the distinction right vs. left is not very important. For one thing, you can consider that right actions are actually left actions, but using a more complicated product, $\star$. As far as group theory is concerned, the actual form of the product is irrelevant, so the theory with $\circ$ and the theory with $\star$ are qualitatively identical. It's just a change of symbol. – AccidentalFourierTransform May 10 '18 at 13:35

I think an "active" transformation is a left action, and a "passive" transformation is a right action.

Active transformation:

$$(R_1 \cdot (R_2 \cdot \phi)) (x) = (R_2 \cdot \phi) (R_1^{-1} x) = \phi(R_2^{-1} R_1^{-1} x) = ((R_1 R_2) \cdot \phi) (x)$$

Passive transformation:

$$(R_1 \cdot (R_2 \cdot \phi)) (x) = (R_2 \cdot \phi) (R_1 x) = \phi(R_2 R_1 x) = ((R_2 R_1) \cdot \phi) (x)$$

Usually physicists don't ever think about right actions, but mathematically there's no reason not to.

• I just wanted to say I feel a little bit bad about not awarding the bounty to you since you identified the problem first, but the other answer is just much more complete. Still, thanks a lot! – Javier May 10 '18 at 14:50