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In $SU(N)$, the set of matrices in the fundamental representation plus the identity:

$$ \left\{ \mathbf{1}, t^a \right\} $$

acts as a basis for generator products under matrix multiplication, such that any combination of the elements in the set can be written as a linear combination, i.e. for any

$$ X_F = t^{a_1}\dotsm t^{a_n} $$

we can write

$$ X_F = c^0 \mathbf{1} + c^a t^a \, .$$

It is easy to show that

$$ c^0 = \frac{1}{N} \text{tr} (X_F) \, , \quad c^a = - 2 i\, \text{tr} (X_F t^a)\, . $$

We cannot do the same for the adjoint representation because the set

$$ \left\{ \mathbf{1}, T^a \right\} $$

doesn't span the full product space, e.g. $T^a T^b$ cannot be expressed solely in terms of $\mathbf{1}$ and $T^c$.

So my question is: can we add - and if yes which - a minimal number of $n^2\!-\!1$ matrices $\{K^a\}$ to $ \left\{ \mathbf{1}, T^a \right\} $ in order to make it a closed set?

EDIT

Where it comes from is that because in the fundamental representation we have additional anti commutation rules

$$ \left\{ t^a, t^b \right\} = \delta^{ab} \frac{\mathbf{1}}{N} + d^{abc} t^c \, , $$

we can take the mean of these and the standard commutation rules to write

$$ t^a t^b = \frac{1}{2} \delta^{ab} \frac{\mathbf{1}}{N} +\frac{1}{2} h^{abc} t^c \,,$$

where $h^{abc}=d^{abc}+i f^{abc}$. We chain this relation to write for any product of fundamental generators

$$ t^{a_1}\cdots t^{a_n} = A^{a_1\cdots a_n} \frac{\mathbf{1}}{N} + B^{a_1\cdots a_n b} t^b \, ,$$

with $A$ and $B$ some constants. This is the same statement as above.

However we cannot do the same in the adjoint representation, because then the anti commutation rules don't close, i.e. they cannot be written as a linear combination of $\mathbf{1}$ and $T^a$.

So the question is, can we add some extra matrices $\{K^a\}$ such that we can write the anti commutation rules as

$$ \left\{ T^a, T^b \right\} = \delta^{ab} \frac{\mathbf{1}}{N} + d^{abc} T^c + c^{abc}K^c\, , $$

which would give

$$ T^{a_1}\cdots T^{a_n} = A^{a_1\cdots a_n} \frac{\mathbf{1}}{N} + B^{a_1\cdots a_n b} T^b + C^{a_1\cdots a_n b} K^c\, .$$

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  • $\begingroup$ As finite-dimensional Lie groups are matrix groups, it is a trivial fact that they (and their Lie algebras) embed into the ring $\mathrm{Mat}_\mathbb{C}(N,N)$ for some $N$, which, as a ring, is closed under multiplication and addition, so you may simply add the matrices of that ring that are independent (as $\mathbb{R}$-vectors) of the algebra elements to the generators to get your "closed set". I don't see the value in that, though. $\endgroup$ – ACuriousMind Sep 5 '14 at 14:54
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    $\begingroup$ You might also be interested in the concept of the universal enveloping algebra $\endgroup$ – ACuriousMind Sep 5 '14 at 15:02
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It seem that what you are seeking is the universal enveloping algebra of a Lie algebra:

Let $\mathfrak{g}$ be our Lie algebra (not necessarily finite dimensional). Now, take the tensor algebra of it as one can do for vector spaces, i.e. $T(\mathfrak{g}) := \bigoplus_{n \in \mathbb{N}}\mathfrak{g}^{\otimes n}$, and divide out the ideal $I_\mathrm{brac}$ generated by $a \otimes b - b \otimes a - [a,b]$, with $[\dot{},\dot{}]$ as the Lie bracket. Dividing out this ideal means imposing the relation $$a \otimes b - b \otimes a = [a,b]$$

onto the otherwise free algebra that is the tensor algebra. Here, we really just declare that we would like to identiy the Lie bracket with the commutator of the tensor algebra. (Note that every associative algebra becomes a Lie algebra if the take the commutator as Lie bracket) Quite naturally, it follows that the inclusion $\iota : \mathfrak{g} \to T(\mathfrak{g})$ descends to a Lie algebra inclusion $\iota' : \mathfrak{g} \to T(\mathfrak{g})/I_\mathrm{brac}$, which is still injective since $I_\mathrm{brac}$ only identifies tensors of the second or higher degree, and $\mathfrak{g}$ embeds as tensors of first degree.

Therefore, $T(\mathfrak{g})/I_\mathrm{brac}$ is the universal enveloping algebra of $\mathfrak{g}$, within which every product of elements of $\mathfrak{g}$ lies.

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The matrices $$\{ 1,t^a \}$$ are the basis vectors of the adjoint representation, not the fundamental representation, because they are matrices. The fundamental representation is the vector space of the column vectors, not matrices!

Your argument that the matrices above do not form the full basis of the adjoint representation of the Lie algebra isn't correct. There is no need why the space spanned by the basis vectors should be closed under multiplication. Instead, because it is Lie algebra, it is closed under commutator and you may indeed verify that the commutator of $1$ with everything else is zero which is OK and the commutators $$ [t^a,t^b] = f^{ab}{}_c t^c $$ are linear combinations of other generators i.e. basis vectors. The coefficients $f$ are known as the structure constants. Let me emphasize that the closure under the commutator is only needed for the vector space or representation to be a Lie algebra; it is not needed for the basis to be a basis of a linear space.

If you extended the space by adding all the products of the matrices and further products etc. you would end up with the space of all matrices of the same size i.e. the Lie algebra of $GL(n)$.

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