One can show that the adjoint representation of $\mathrm{SU}(n)$, the image of the map $\mathrm{Ad}:\mathrm{SU}(n) \rightarrow \mathrm{Aut}(\mathrm{su(n)})\subset \mathrm{GL}(\mathrm{su}(n))$, is an $n^2-1$-dimensional lie subgroup of $\mathrm{SO}(n^2-1)$.

I would to know if any characterisations of this subgroup are known? In particular, I would like to understand what object, apart from the inner product, is preserved under the action of elements from the lie group $\mathrm{Im}(\mathrm{Ad})\subset \mathrm{SO}(n^2-1)$. Alternatively, I am looking to understand the action of $\mathrm{Im}(\mathrm{Ad})$ on $\mathbb{R}^{n^2-1}$.

I know that given a basis for $\mathrm{su}(n)$ one can construct a basis for the Lie algebra of $\mathrm{Im}(\mathrm{Ad})$ using the $\mathrm{ad}$ map and the structure constants for $\mathrm{su}(n)$ - However I cannot see how to use this to answer the question above?

In addition, I am aware that $\mathrm{Im}(\mathrm{Ad}) \simeq \mathrm{SU}(n)/\mathbb{Z}_n$ - However I also cannot see how to use this to help me answer the above question? In the $n=2$ case one can use the first isomorphism theorem along with the well know group homomorphism between $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$, however for the general case it seems like this approach would require the construction of a group homomorphism $f:\mathrm{SU}(n)\rightarrow \mathrm{SO}(n^2-1)$ with kernel $\mathbb{Z}_n$? It seems unlikely one can construct such a homomorphism in general without the useful parameterizations of $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ one has for the case where $n=2$? Is there another way to exploit this quotient group structure?

  • 2
    $\begingroup$ Actually, your first sentence contains an error. The dimension of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ is $n^2{-}1$, not $n$. $\endgroup$ – Robert Bryant Aug 26 '14 at 20:49

Here is a different characterization of the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{SO}(n^2{-}1)$ that works when $n>2$.

Define a skew-symmetric trilinear form $\kappa:{\frak{su}}(n)\times{\frak{su}}(n)\times{\frak{su}}(n)\to\mathbb{R}$ by the formula $$ \kappa(x,y,z)=\mathrm{tr}(xyz-xzy)\ \bigl(=\mathrm{tr}(yzx-yxz)=\mathrm{tr}(zxy-zyx)\bigr). $$ Then, for $n>2$, the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ is the identity component of the subgroup of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ that preserves $\kappa$. (This doesn't work for $n=2$, of course. Also, you do have to restrict to the identity component because $\mathrm{SU}(n)$ has an outer automorphism (namely, conjugation) when $n>2$, and this preserves $\kappa$ as well; accordingly, the full $\kappa$-stabilizer has two components.)

In a certain sense, this is the 'simplest' characterization of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as a subgroup of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$. What I mean is that, if you want to describe this subgroup as the set of linear transformations that preserve some set of algebraic objects on the vector space ${\frak{su}}(n)$, then $\kappa$ is the object that sits in the smallest representation of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ that does the job.

For example, if you want to define $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup that preserves the Lie bracket $[,]:{\frak{su}}(n)\times {\frak{su}}(n)\to {\frak{su}}(n)$, then you have to think of $[,]$ as an element of ${\frak{su}}(n)\otimes {\frak{su}}(n)^\ast\otimes {\frak{su}}(n)^\ast$ or, if you want to build in the skew-symmetry of the bracket, as an element of ${\frak{su}}(n)\otimes \Lambda^2({\frak{su}}(n)^\ast)$. Either way, these vector spaces have a much higher dimension than $\Lambda^3({\frak{su}}(n)^\ast)$, which is where $\kappa$ lives.

As another example, if you want to think of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup that preserves the quadratic form $Q(x) = \mathrm{tr}(x^2)$ and the ($\mathbb{R}$-valued) cubic form $C(x) = i\,\mathrm{tr}(x^3)$, you are asking that it preserve an element of $S^2({\frak{su}}(n)^\ast)\oplus S^3({\frak{su}}(n)^\ast)$, and, again, this has considerably higher dimension than $\Lambda^3({\frak{su}}(n)^\ast)$. You can cut this down a little bit by noting that $C$ is harmonic with respect to $Q$, so that you could look at characterizing $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup of $\mathrm{O}\bigl({\frak{su}}(n),Q\bigr)$ that stabilizes $C\in S^3_0({\frak{su}}(n)^\ast,Q)$, but this still is more total equations than the equations that say that an element fixes $\kappa$.

I'm not saying that using $\kappa$ to define $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ is the most useful way to define it, I'm just saying that it's algebraically the least redundant.

For example, look at the case $n=3$: The codimension of $\mathrm{Ad}\bigl(\mathrm{SU}(3)\bigr)$ in $\mathrm{GL}\bigl({\frak{su}}(3)\bigr)\simeq \mathrm{GL}(8,\mathbb{R})$ is $64-8 = 56$, which is equal to the dimension of $\Lambda^3\bigl(\mathbb{R}^8\bigr)$, so it follows that the $\mathrm{GL}(8,\mathbb{R})$-orbit of $\kappa$ is actually open in $\Lambda^3\bigl(\mathbb{R}^8\bigr)$. Thus, the condition of fixing $\kappa$ is $56$ independent equations on an element of $\mathrm{GL}(8,\mathbb{R})$, which is the absolute minimum number of equations you need to cut out $\mathrm{Ad}\bigl(\mathrm{SU}(3)\bigr)$ as a submanifold.


As your question already suggests, the "extra object preserved" is the Lie bracket. I.e., for any connected semisimple Lie group $G$, the image of $\mathrm{Ad}: G\to\mathrm{GL}(\mathfrak g)$ is precisely the identity component $\mathrm{Aut}(\mathfrak g)^0$ of $$ \mathrm{Aut}(\mathfrak g) = \{a\in\mathrm{GL}(\mathfrak g):a([X,Y])=[a(X),a(Y)] \text{ for all } X,Y\in\mathfrak g\}. $$

Proof. Clearly $\mathrm{Ad}$ maps $G$ into that identity component. To verify that it is onto it, it is enough to see that its derivative $\mathrm{ad}:\mathfrak{g}\to\mathrm{gl}(\mathfrak g)$ is onto the Lie algebra $$ \mathrm{aut}(\mathfrak g) = \{D\in\mathrm{gl}(\mathfrak g):D([X,Y])=[D(X),Y]+[X,D(Y)] \text{ for all } X,Y\in\mathfrak g\} $$ of $\mathrm{Aut}(\mathfrak g)$, or in other words, that every derivation $D$ of $\mathfrak g$ is inner: $D=\mathrm{ad}(Z)$ for some $Z\in\mathfrak g$. But that is precisely the content of Whitehead's Lemma (see any book on semisimple Lie algebras).

  • $\begingroup$ Thanks, this is useful - I will have to put some thought into how to use this preservation of the lie-bracket to understand the action of a generic element of $\mathrm{Aut}(\mathrm{su}(n))$ - which is also an element of $\mathrm{SO}(n^2-1)$ - on $\mathbb{R}^{n^2-1}$. $\endgroup$ – Ryan Aug 21 '14 at 10:39

I am not sure if I am providing a characterization of the sort you seek. Nevertheless, the Lie group $SU(n)/\mathbb{Z}_n$ is called the projective special unitary group, $PSU(n)$. Note that this group has a natural action on $\mathbb{P}^{n-1}$. In fact, it is precisely the group of holomorphic isometries of $\mathbb{P}^{n-1}$.

  • $\begingroup$ Thank you - do you know of a good reference I could use to read up on this group and its action? In particular I am not sure of the definition of a holomorphic isometry? $\endgroup$ – Ryan Aug 21 '14 at 9:39
  • $\begingroup$ The usual Hermitian inner product on $\mathbb{C}^n$ yields a Riemannian metric on $\mathbb{P}^{n-1}$. A holomorphic isometry of $\mathbb{P}^{n-1}$ is a biholomorphism of $\mathbb{P}^{n-1}$ with itself that preserves this metric. One quick reference for the group, its action on $\mathbb{P}^{n-1}$, and the metric is Einstein Manifolds by Arthur Besse, pages 178-181. $\endgroup$ – Peter Crooks Aug 21 '14 at 9:52
  • $\begingroup$ Thanks, I will have a look at this reference. However my initial impression is that this is not quite the characterisation I am looking for - I will investigate none the less! $\endgroup$ – Ryan Aug 21 '14 at 10:01

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