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while reading Freedman & van Proeyen's book, I found a very strange claim concerning representation on Lie Algebras: they define a generic transformation spanned by the parameter $ε^A$ and "abstract" element of the algebra $T_A$ $$ δ(ε)=ε^AT_A,\tag{1} $$ then on fields $$ T_Aφ^i = -(t_A)^i_jφ^j\tag{2} $$ where $t_A$ is a matrix. Now they claim that

... a symmetry transformation acts on the field, which are dynamical variables of the system, and not of the matrices, which are the result of a prior transformation.

to be more explicit they write: $$ \begin{align} δ(ε_1)δ(ε_2)φ^i & = ε_1^AT_Aε_2^B\left[-(t_B)^i_jφ^j \right]\\ & = ε^A_1ε^B_2(-t_B)^i_jT_Aφ^j\\ & = ε^A_1ε^B_2(-t_B)^i_j(-t_A)^j_kφ^k. \end{align} $$

That seems a right action since the order of matrices $t$ is inverted respect to the abstract element of Lie Algebra $T$, but the matrices are composed as an (ordinary) left action! They further comment on that:

It is important to realize that, in the second line, the transformation operator acts on the field $φ^j$ , and not on the ‘numbers’ $(t_B )^i_j$.

This claim and the whole motivation above seems to me at leas very awkward since to me $(t_A)^i_jφ^j$ (sum on $j$) is itself a field. Not just $φ^j$ but also when you multiply it by the matrix $t_A$. This is for me rather obvious since one can define $$ φ'^i:=(t_A)^j_iφ^j $$ and so the "field detector" fails. How can $T_A$ know if the field on which it acts has been previously acted on by other $T$s? Indeed when I perform a field transformation I expect to obtain again a field as a result, a field with the same dignity of the first, with the same properties, and therefore that cannot be discriminated from the first one. How can they give a motivation like that? Is it mathematically correct? (I don't think so)

WAY OUT

In my opinion the correct definition to have a right action in stead of (1) is $$ T_Aφ^i=φ^j(-t_A)^i_j $$ so that (2) becomes $$ \begin{align} δ(ε_1)δ(ε_2)φ^i & = ε_1^AT_Aε_2^B\left[φ^j(-t_B)_j^i \right]\\ &= ε_1^Aε_2^Bφ^k(-t_B)_k^j(-t_A)_j^i \end{align} $$

The point is that this cannot be a simple type since they motivate their passages with that weird claim above.

What do you think about?

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  • $\begingroup$ Review your passive transformations. Consider a rotation of a vector, so components dotted on basis vectors. The components are "numbers" in this language. Rotate the basis vectors and behold the right action on the rotation generators. Do you now see the significance of the - sign? Try an obvious example. Compose, etc. $\endgroup$ – Cosmas Zachos Feb 11 '17 at 12:38
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Let us first properly define all objects in play here. Let $\mathfrak{g}$ be the Lie algebra with chosen basis $T_A$ and $V$ the target space of the fields. We have maps $$ \delta : \mathbb{R}^{\dim(\mathfrak{g})}\to \mathfrak{g},\epsilon\mapsto \epsilon^A T_A$$ and $$ \rho : \mathfrak{g}\to \mathfrak{gl}(V), T_A\mapsto t_A.$$

$\delta$ is compatible with the Lie bracket in the sense that $$ [\delta(\epsilon_1),\delta(\epsilon_2)] = \epsilon^A_1\epsilon^B_2[T_A,T_B]$$ but we may notice that the definition of $\rho$ is incomplete: It is defined on the basis and linearly extended but we don't know how it behaves with respect to the Lie algebra structure. In particular, we don't know whether $\rho([T_A,T_B]) = [t_A,t_B]$ or not. Note that asking what $\rho(T_AT_B)$ is non-sensical because this product is not defined on the level of the abstract Lie algebra itself, so we really have no a priori expectations about what the $\rho(T_A)\rho(T_B)$ should be equal to.

All the rather confusing "arguments" presented in your question essentially amount to defining $\rho$ to fulfill $\rho([T_A,T_B]) = [t_B,t_A]$. For an example of such a map, consider the concatenation of a usual representation of a Lie algebra $\sigma$ with $\sigma([T_A,T_B]) = [\sigma(T_A),\sigma(T_B)]$ with the transpose or the adjoint. You are correct that this gives formally a right action of the (enveloping algebra of the) Lie algebra.

As Cosmas Zachos comments, the true underlying motivation is that your source seems to consider the action of the Lie algebra as a passive transformation, which acts on the basis and not on the vectors.

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  • $\begingroup$ $\mathfrak{gl}(V)$ is a well-known Lie algebra which does also have an associative product, namely composition. So $\rho$ is perfectly defined and $\rho(T_A) \rho(T_B)$ makes sense. It is then true that $\rho$ may not be a Lie algebra morphism at all if one chooses the $t_a$ arbitrarily. In particular, having $\rho\big( [T_A, T_B]\big)= [t_B, t_A]= \left[-\rho(T_B), -\rho(T_A) \right] = \left[\rho(T_B), \rho(T_A) \right] $ says that $\rho$ is not a Lie algebra morphism (wrong order), unless the commutation are actually 0. $\endgroup$ – Noix07 May 28 '17 at 20:19
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There is indeed a problem with the definition of your book. As you have already written, if one is given $\varphi'\,^i = (t_A)^i{}_j\, \varphi^j$ without knowing that this field was already obtained from a previous transformation (i.e. without knowing the r.h.s.) then how should it be transformed?

One can check that decomposing $\varphi'$ in different ways will lead to different transformations: $$\left\lbrace \begin{aligned} \text{if}\ \varphi'\,^i = M^i{}_j\, \varphi^j\ \text{then }\ T_A\, \varphi'\, ^i &= - M^i{}_j\,(t_A)^j{}_k\, \varphi^k\\ \text{but if}\ \varphi'\,^i = \varphi^i\ \text{then }\ T_A\, \varphi'\, ^ i &= - (t_A)^i{}_j\, \varphi^j \end{aligned} \right. $$

Also the authors obviously confuse the transformation of the group and that of the Lie algebra: schematically $\ g = 1 + \epsilon^A \, T_A + o(\epsilon)$ in the group so that $$ \begin{equation} \begin{split} g_1\cdot g_2 &= (1 + \epsilon_1^A \, T_A) \cdot (1 + \epsilon_2^A \, T_A) + o(\epsilon_1, \epsilon_2)\\ &= 1 + \epsilon_1^A \, T_A + \epsilon_2^A \, T_A + o(\epsilon_1, \epsilon_2) \end{split} \end{equation}$$ so that the successive action of the algebra should correspond to adding the actions (vs. composition for the group).

Finally if one tries the "passive" transformation (for a matrix group if one is to calculate) one indeed finds a right action, simply because one takes inverse: $$g\cdot \varphi = (M_g^{-1})^i{}_j \, \varphi^j$$ where $g$ denotes an element in the group and $M_g$ the associated matrix. Then for the algebra, if $M_g= \mathbb{1} + \epsilon^A\, T_A + o(\epsilon)$ then $M_g^{-1}= \mathbb{1} - \epsilon^A\, T_A + o(\epsilon)$ which explains the minus in $ T_A\, \varphi' = - (t_A)^j{}_k\, \varphi^k$

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