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At least for low values of $N$ like $2$ or $3$ and such I would like to know if there are explicit matrices known giving the representation of $u(N)$ or $U(N)$ in the adjoint?

(..a related query: Is it for the Lie group or the Lie algebra of U(N) that it is true that the weight vectors in the fundamental/vector representation can be taken to be N N-vectors such that all have weight/eigenvalue 1 under its Cartan and the ith of them has 1 in the ith place and 0 elsewhere and for the conjugate of the above representation its the same but now with (-1)?..I guess its for the u(N) since they are skew-Hermitian but would still like to know of a precise answer/proof..)

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I've only recently started learning about group theory so it would be worth double checking the following.

As I understand it, when we say something (for example a field, $X$) transforms under the adjoint rep of a group (say $U(N)$), the transformation can involve any matrix, $g$, belonging to the group $U(N)$. The adjoint rep doesn't put any constraints on the matrices we choose from $U(N)$, rather it tells us how to use those matrices in a transformation. When we say $X$ transforms in the adjoint rep of $U(N)$ it simply means it transforms as $X \rightarrow X' = gXg^{-1}$. The matrix $g$ can be any matrix belonging to $U(N)$. Note: $g^{-1}$ is the inverse of the matrix $g$. So when we say 'an object is in the adjoint representation of $U(N)$' this is a statement about how to act the matrices of $U(N)$ on an that object. It doesn't say anything about which matrices we choose from $U(N)$.

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    $\begingroup$ Criticism/ confirmation is more than welcome from other users with regards to this answer. $\endgroup$ – user15766 May 24 '13 at 10:12
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    $\begingroup$ By the sounds of it, you are a maths-physics student too, since you say "transforms adjointly". I have only noticed other maths-physics saying that, the pure maths guys do not seem to. Anyway, I have wondered about this before. I think it is the case that when we say "X transforms in the adjoint rep of G" what we mean is "the action of the group G on X is the adjoint group action". I don't know if it's always the adjoint representation. Mostly because sometimes I notice the dimensions do not equat correctly for it to be the adjoint representation. I am not so sure either though. $\endgroup$ – Flint72 May 14 '14 at 21:05
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    $\begingroup$ @user15733 : as regards your statement "The adjoint rep doesn't put any constraints on the matrices we choose from [G]", this is incorrect. The adjoint representation is defined as acting on a vector space of equal dimension to the dimension of G, so $N^2$ in the case of $U(N)$. I think you are getting confused between the idea of the adjoint representation, and the adjoint group action. $\endgroup$ – Flint72 May 14 '14 at 21:07
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I don't know if it helps, but I think you can write $$U(N)=SU(N)\times U(1)$$ usually (in physics) people consider semisimple groups, i.e. groups without U(1) factors (=tori). To give an explicit example, consider $U(2)=SU(2)\times U(1)$. The adjoint representation of the algebra can be defined by $$(T_a)_{bc}=-if_{abc}$$ where $f_{abc}$ are the structure constants of the algebra. Perfoming an explicit calculation for $SU(2)\times U(1)$ and keeping in mind that the generator of $U(1)$ commutes with all other elements of the algebra we get the following generators for the adjoint representation: $$ T_1=\begin{pmatrix}0&0&0&0\\0&0&-i&0\\0&i&0&0\\0&0&0&0\end{pmatrix} T_2=\begin{pmatrix}0&0&i&0\\0&0&0&0\\-i&0&0&0\\0&0&0&0\end{pmatrix} T_3=\begin{pmatrix}0&-i&0&0\\i&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix} T_4=0 $$ So you see, that with $T_1, T_2, T_3$ you can span a 3 dimensional real vector space, but with all four you can't since $T_4$ vanishes. This means, as I think, that while you can define an adjoint representation for $SU(2)$, you can't do so for $U(2)$ or $U(1)$.

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  • $\begingroup$ No, you cannot write that! You need a semidirect product. $\endgroup$ – Benjamin Jun 5 '15 at 13:50
  • $\begingroup$ Hi Benjamin, thanks for the comment. Could you explain what the right expression for would be, so I can edit my answer? How does this affect the algebra? $\endgroup$ – Stan Jun 8 '15 at 9:22
  • $\begingroup$ A clear explanation can be found here en.m.wikipedia.org/wiki/Unitary_group in the properties section. $\endgroup$ – Benjamin Jun 8 '15 at 9:55
  • $\begingroup$ Hi Benjamin. Uhm, well, there's not much more than what you've already mentioned, namely that $U(1)\lhd SU(N)$, and therefore $U(N) = SU(N)\ltimes U(1)$. But how does this affect the algebra and the adjoint representation? $\endgroup$ – Stan Jun 8 '15 at 11:43
  • $\begingroup$ No, that's not what's there! $U(n)$ is a semidirect product of $SU(n)$ and $U(1)$. Try it for yourself now, write down a general curve through the identity and differentiate. This will find what the algebra is in both cases. As for the adjount rep, you will see straight off after finding what the algebras are. Good luck. $\endgroup$ – Benjamin Jun 8 '15 at 15:35

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