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I'm following notes that define the adjoint representation as a map $\rho$ from $\mathfrak{g}$ to $End(\mathfrak{g})$ such that, for $t_1, t_2$ $\in \mathfrak{g}$ ,

\begin{equation} \rho_{\tiny{adj}}(t_1) (t_2) = [t_1,t_2]. \end{equation}

Taking a basis $t_\alpha$ for $\mathfrak{g}$ we know that

\begin{equation} [t_\alpha,t_\beta]=f_{\alpha\beta}^{\,\,\,\,\gamma} t_\gamma. \end{equation} How do we see that this matches the alternative definition

\begin{equation} \rho_{\tiny{adj}}(t_\alpha)_{\beta}^{\,\,\,\gamma} = f_{\alpha\beta}^{\,\,\,\,\gamma}. \end{equation}


Attempt: \begin{equation} \rho(t_\alpha) (t_\beta) = [t_\alpha,t_\beta] = f_{\alpha\beta}^{\,\,\,\,\gamma} t_\gamma. \end{equation} However I seem to get confused with the indices because it seems impossible to me to get $\rho(t_\alpha)$ to be $f_{\alpha\beta}^{\,\,\,\,\gamma}$. Writing the indices explicitly I would think we get

\begin{equation} \rho(t_\alpha)_{\mu}^{\,\,\,\nu} (t_\beta)_{\nu} = [t_\alpha,t_\beta]_{\mu} = f_{\alpha\beta}^{\,\,\,\,\gamma} (t_\gamma)_{\mu}. \end{equation}

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  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Jan 20 '17 at 12:32
  • $\begingroup$ @Qmechanic. Maybe, I did consider that but I would prefer a physicist's answer as they are less likely to contain mathematics that I don't know, not being a mathematician myself. If this should be moved that is okay with me too. (Is there a good way to do that myself? Or should someone else do that?). $\endgroup$ – Kvothe Jan 20 '17 at 12:54
  • $\begingroup$ @Kvothe: What notes/book are you using? $\endgroup$ – CAF Jan 20 '17 at 21:53
  • $\begingroup$ @CAF. Notes by Matthias Gaberdiel edu.itp.phys.ethz.ch/hs13/Symmetries/notes/Symmetries.pdf Gabradiel gives the above definition in 3.1.19 on p.40 and one page later in 3.2.7 he uses the fact that the adjoint representation is given by the structure constants without proving this. (The notes are very similar to the standard Fulton &Harris book.) $\endgroup$ – Kvothe Jan 21 '17 at 22:24
  • $\begingroup$ @Kvothe: See my answer below for my take on things :) $\endgroup$ – CAF Jan 22 '17 at 11:56
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The adjoint map at the level of the lie algebra is such that $$\text{ad}: \mathfrak{g} \rightarrow \text{End}(\mathfrak g),$$ taking $\lambda_a \mapsto \text{ad}_{\lambda_a}$ where $$\text{ad}_{\lambda_a}: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\, \lambda_b \mapsto [\lambda_a, \lambda_b].$$

So, this means that $$\lambda_b \rightarrow \text{ad}_{\lambda_a} \lambda_b = [\lambda_a, \lambda_b] = if_{abc}\lambda_c.$$

The map $\text{ad}$ is equivalent to $\rho$ for matrix lie groups so $\rho: \mathfrak{g} \rightarrow \mathfrak{gl}(V=\mathfrak{g}) \equiv \text{End}(V = \mathfrak{g}) $ is the map such that $\rho(\lambda_a)_{cb}\lambda_c = [\lambda_a, \lambda_b] = if_{abc}\lambda_c$, which is to say $\rho(\lambda_a) (\lambda_b) = [\lambda_a, \lambda_b].$ Peeling off the $\lambda_c$ we obtain $$\rho(\lambda_a)_{cb} = i f_{abc}.$$

So the indices $\left\{b,c\right\} \in \left\{1, \dots , \text{dim}(\mathfrak{g})\right\}$ attached to the $\text{dim}(\mathfrak{g}) \times \text{dim}(\mathfrak{g})$ matrix representations $\rho(\lambda_a)_{bc}$ of the lie algebra adjoint action mix around the real $\text{dim}(\mathfrak{g}) \times 1$ basis vectors $\left\{\lambda \right\}$ of the lie algebra.

Alternatively, since $\rho$ is a linear operator acting on the vector space $V = \mathfrak{g}$, an equivalent operation is one in which $\rho$ mixes around the components $(\lambda_b)_d$ of some $\lambda_b$, so we can equally rewrite the transformation law $$\rho(\lambda_a)_{cd} (\lambda_b)_d = \rho(\lambda_a)_{cd} \delta_{bd} = \rho(\lambda_a)_{cb} \overset{!}{=} [\lambda_a, \lambda_b]_c = if_{abd} (\lambda_d)_c = if_{abd} \delta_{dc} = if_{abc}$$ and thus arrive at the same conclusion $$\rho(\lambda_a)_{cb} = if_{abc},$$ where $(\lambda_i)_j = \delta_{ij}$ WLOG (if there is an overall normalisation, this cancels between the two sides of the equation) and $|\lambda_b \rangle = (\lambda_b)_d |\lambda_d \rangle$.

$\mathbf{Aside}$

One may ask why $\rho(\lambda_a)_{cb} \lambda_c$ and not, say, $\rho(\lambda_a)_{bc} \lambda_c$? We define the action of a generator $\lambda_a$ on a basis vector $|\lambda_b \rangle$ through the Lie bracket as noted above. This is to say $\lambda_a |\lambda_b \rangle = \rho(\lambda_a)_{cb} |\lambda_c \rangle$ so that $\langle \lambda_c | \lambda_a |\lambda_b \rangle = \rho(\lambda_a)_{cb}$ are the matrix elements of the matrix representation of the operator $\rho$ in the basis $\left\{\lambda\right\}$. Consider some arbitrary vector built up from the basis vectors of $\mathfrak{g}$, that is $|\Psi \rangle = \sum c_b |\lambda_b \rangle$. Then $$|\Psi \rangle' = c_d' |\lambda_d \rangle \equiv \lambda_a |\Psi \rangle = c_b \lambda_a |\lambda_b \rangle = c_b \rho(\lambda_a)_{cb} |\lambda_c \rangle $$ Relabelling indices at the first equality allows us to write the transformation of the components of the vector $|\Psi \rangle$ in the basis spanned by $\left\{\lambda \right\}$ as $c_c' = \rho_{cb} c_b$.

All of this makes contact with the familiar matrix/vector product $\mathbf v' = M \mathbf v$ or in component form $v_i' = M_{ij}v_j$ and $\mathbf v = v_i e_i$ where $v_i$ are the components of $\mathbf v$ in the basis $\left\{e_j \right\}$.

In the notation of the aside, the $\rho(\lambda_a)$, $|\Psi \rangle, |\lambda_b \rangle$ and $c_b$ would be equivalent to the $M$, $\mathbf v$, $e_j$ and $v_j$ above respectively. So, we see that with the used indicial writing of the $\rho$ we obtain the familiar transformation law for components of vectors.

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  • $\begingroup$ Thanks @CAF, I agree for the msot part. Still, I'm a bit uneasy about getting $[\lambda_a, \lambda_b]$ and not something with $[\lambda_a, \lambda_c]$. (although at that point there is nothing you can do anymore to make the indices work out alright). I guess it comes down to feeling wrong about $\rho(\lambda_a) (\lambda_b)$ being $\rho(\lambda_a)_{bc}\lambda_c $. How did the b get down there? Of course $\rho(\lambda_a)$ is something that mixes up the $\lambda$'s and it does this in a b (and a) dependent way. It is clear that this is the only way it could possibly act. But still... $\endgroup$ – Kvothe Jan 22 '17 at 17:27
  • $\begingroup$ The thing is that usually with for example a vector V we would have M V or in index notation $M_{ab} V_b$. Here we have $\rho_{\lambda_a}$ working on a vector $\lambda_b \in \mathfrak{g}$ labelled with a $b$ to indicate which element - or base vector - we have. So this index $b$ can equivalently be seen as its component index? As far as only the indices tell me there is one other structure that could be meant by $\rho(\lambda_a) (\lambda_b)$ and that is $\rho(\lambda_a)_{cb}\lambda_c$, this would differ by a sign. How can we check that this is not correct? $\endgroup$ – Kvothe Jan 23 '17 at 8:56
  • $\begingroup$ My replies with @ seem to vanish. Very strange. Hope this reaches you. Or is this always true that you don't see references you made yourself. $\endgroup$ – Kvothe Jan 23 '17 at 9:01
  • $\begingroup$ To answer my own last (meta) question: "The auto-complete box can determine when an @name would not be necessary to trigger a notification and in that case will not include the user in the list." Also luckily my first reply with @ was unnecessary as I screwed it up. @ needs to be at the start of the comment. (Confusing me even more by not vanishing there. It didn't vanish there because while it was not necessary it also was invalid). $\endgroup$ – Kvothe Jan 23 '17 at 9:12
  • $\begingroup$ The latter statement I suppose can be seen by considering such a rewriting for the defining equation for lorentz transformation: $\eta_{\rho \sigma} = \eta_{\mu \nu} \Lambda^{\mu}_{\,\,\rho} \Lambda^{\nu}_{\,\,\sigma}$ which is not in matrix notation just $\eta \Lambda \Lambda$ or $\Lambda \eta \Lambda$. Here one has to match up all the indices correctly and write $\eta_{\rho \sigma} = (\Lambda^T)_{\rho}^{\,\,\mu}\eta_{\mu \nu} \Lambda^{\nu}_{\,\,\sigma}$ which means $\eta = \Lambda^T \eta \Lambda$ in matrix notation :) $\endgroup$ – CAF Jan 24 '17 at 9:44

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