# Why is the Lie algebra of the unitary group generated by matrix units?

I have read this specifically in chapter 2.10 of the book Lie algebras and applications by Francesco Iachello, but I also saw similar definitions in several physicist papers: The Lie algebra of $$\mathrm{U}(d)$$ is generated by the matrix units $$E_{ij}:=|j\rangle\langle i|$$, where $$\{|i\rangle\}_{i=1}^N$$ is an orthonormal basis on $$\mathbb{C}^d$$. Consequently, we can obtain a set of generators for the Lie algebra of SU(d), if we make these matrices traceless, i.e., by setting $$\overline{E}_{i,j}:=|j\rangle\langle i|-\frac{1}{d}\delta_{i,j}\mathrm{Id}$$. (Only $$d^2-1$$ of these generators are independent.)

As far as I understand, the Lie algebra of $$\mathrm{U}(d)$$ should contain only skew-hermitian matrices, and these generators obviously aren't that. For example, $$\exp(E_{1d})$$ is not unitary, when it should be if $$E_{1d}$$ really was an element of the Lie algebra. In fact, I believe that the Lie algebra generated by the set $$\{E_{i,j}\}_{i,j=1}^d$$ is actually that of the general linear group. What am I misunderstanding here?

• $E_{ij}$ is not hermitian (or anti hermitian) so its exponential is not in $U(n)$. You need to exponentiate hermitian combos of $E$’s to get an element in $U(n)$. The $E$’s do form a basis for the algebra.. – ZeroTheHero Mar 18 at 21:29

You are certainly correct that the Lie algebra of $$U(d)$$ consists of skew-Hermitian $$d \times d$$ matrices. However, physicists will often implicitly complexify Lie algebras, without ever bothering to mention that they are doing it. The complexification of $$u(d)$$ is indeed $$gl(d, \mathbb{C})$$. That's because multiplying by $$i$$ we get Hermitian matrix, and any matrix at all can be expressed as the sum of a Hermitian and a skew-Hermitian matrix. You can find more discussions about Lie groups and algebras in mathematics vs. physics in Peter Woit's book 'Quantum Theory, Groups and Representations'.

• Thanks, This explains my misunderstanding but also begs for a new question: When is it appropriate to consider the complexification of the Lie algebra instead of the original one? The context I read about matrix units as generators was the construction of Casimir operators. If I use the matrix units as the generators to construct Casimir operators, then I actually get Casimir operators of $gl(d,\mathbb{C})$. How do the Casimir operators of $gl(d,\mathbb{C})$ relate to those of $u(d)$? I don't think they necessarily live in the (real) universal enveloping algebra of $u(d)$. – cosmicjoke Mar 19 at 14:59
• I'm not sure; consider asking this as a new question (with more context), preferably over at math stackexchange or mathoverflow. – Pedro Mar 19 at 17:19

As it is often the case the issue lies in conventions. For mathematicians the Lie Algebra generators are any basis that can span the algebra as a vector space, for physicist we usually require the generators themselves to be hermitian (e.g. think about the Pauli matrices), because of their interpretation as observables.

Let us also make clear the difference between Algebra and Group. The Lie algebra is a vector space and an algebra thanks to the Lie bracket (the commutator), it is denoted with small cap fraktur letters. So for this case: $$\mathfrak{u}(N)$$ is the Lie Algebra of the group of unitary matrices of dimension $$N\times N$$. The group is denoted by $$U(N)$$ and is a group under usual matrix multiplication, whose elements are indeed unitary (thus complex entries) matrices.

The algebra condition on the elements of the algebra is obtained by differentiating the unitarity condition, $$A A^\dagger = 1,$$ which gives $$A + A^\dagger = 0$$ which is nothing else than the anti-hermitian condition. This means the Lie algebra is the vector space of all anti-hermitian matrices of dimension $$N\times N$$. So for a matrix $$A\in \mathfrak{u}(N)$$ the exponentiation does give you an element of $$U(N)$$, and it can be shown that all elements in the vicinity of the identity of $$U(N)$$ can be described by exponentiation of some element of $$\mathfrak{u}(N)$$. At this point we are free to describe the matrix $$A$$ as we wish. As physicist we want to pick a basis of hermitian elements (notice that if $$A$$ is anti-hermitian then $$-iA$$ is Hermitian, or vice versa if $$\sigma$$ is Hermitian then $$i\sigma$$ is anti-hermitian) so let us have a basis for $$\mathfrak{u}(N)$$ of hermitian elements multiplied by $$i$$ and voila $$A = \sum_n c_n i \sigma_n \in \mathfrak{u}(N)$$ and exponentiating $$\exp(A) = \exp\left(\sum_n c_n i \sigma_n \right) \in U(N)$$

For the case of $$SU(N)$$ it is the $$\det = 1$$ condition that when differentiated forces the matrices of its algebra to be traceless.

So far we have spoken implicitly of Lie algebras as real vector spaces, that is the $$c_n$$ above are real and don't change the hermitian properties of $$A$$. However one can also complexify the algebra (build a new algebra), so one obtains a vector space over the complex numbers, thus allowing for anti-hermitian and hermitian matrices. So we have \begin{align} \mathfrak{u}(N) &= \text{anti-hermitian matrices}\\[7pt] \mathfrak{u}_\mathbb{C}(N) &= gl(N,\mathbb{C}) = \text{complex matrices size} N\times N \end{align}

• You should write lower case $gl(N, \mathbb{C})$ when you're referring to the algebra of all complex matrices. Upper case is reserved for the group of invertible matrices. – Pedro Mar 18 at 23:43
• Indeed, thank you for the correction. – ohneVal Mar 19 at 7:07

As far as I remember, it is only a matter of definition of the term "generator". If $$H$$ is hermitian $$H^+=H$$ then $$G=iH$$ is skew hermitian: $$H^+=(iG)^+=-iG=-H$$ So after all, the parameter inside the exponential differs only by an imaginary unit, which doesn't change a lot, at least notation-wise. Instead of generating a one-parameter sub-group by $$\exp(\lambda G)$$ where $$G$$ is skew-hermitian, you generate it by $$\exp(i\lambda H)$$ where $$H$$ is hermitian.

• Yes, I understand that whether the elements of the Lie algebra are Hermitian or Skew hermitian depends on whether we define multiplication by i in the exponential map. Physicist tend to use i while mathematicians don't. However, this doesn't really answer the questions because the matrix units are not Hermitian when they are off-diagonal, and $\exp(iE_{ij})$ is not unitary for $i\neq j$. – cosmicjoke Mar 18 at 20:48
• @cosmicjoke: yes, I missed that detail. Indeed, it seems you are right, the algebra basis mentioned in the textbook generates the general linear group on $C^d$. Of course, since $U(d)$ is a subgroup of the general linear group, the general basis is also a basis of the algebra of $U(d)$ (when properly linearly combined...). However, it is a bit strange to express it that way. But the context is missing in your question, so maybe it is explained a bit better in the book. – oliver Mar 18 at 21:43