# What is the dimensionality of each part of a covariant derivative?

In the standard model, we have the following covariant derivative:

$$D_\mu = \partial_\mu - ig_sG_\mu^a\lambda_a-igW_\mu^a\frac{\sigma^a}{2}-ig'B_\mu\frac{Y}{2}$$

If we let this work in on e.g. the lefthanded quark ($$SU(2)$$) doublet $$Q_L$$ then how exactly does this work dimensionally?

$$G$$, $$W$$ and $$B$$ are $$4\times1$$ vectors
$$\sigma^a$$ are $$2\times2$$ matrices
For the first component $$\mu = 0$$, this means that the third term will give us scalar x matrix x doublet = $$2\times1$$.
But $$\lambda^a$$ in the second term is a $$3\times3$$ matrix, so this doesn't work when applying the same logic.

Do I need to find a 2D representation of $$SU(3)$$? Is there even one? I guess you could go to a 3D representation of $$SU(2)$$ instead, but then I don't see how this works on the $$SU(2)$$ doublet.

It's neither $$2 \times 2$$ nor $$3 \times 3$$. It's in the tensor product space, which is much larger.
First consider a simpler example. A single quark is a Dirac spinor in the $$3$$ of $$SU(3)$$. A Dirac spinor has $$4$$ components, but a color triplet has $$3$$ components, so does the quark really have $$3$$ or $$4$$ components? Neither. It actually has $$12$$ components, which might be written as $$q^i_\alpha, \quad i \in \{1, 2, 3\}, \quad \alpha \in \{0, 1, 2, 3\}.$$ All matrices that act on the quark field are $$12 \times 12$$, usually the tensor product of a $$3 \times 3$$ matrix and a $$4 \times 4$$ matrix, so properly we should write something like $$M^{ij}_{\alpha \beta} q^j_\beta.$$ However, writing this many indices gets tiresome, so we suppress them whenever we can, such as if some of the tensor product factors are just the identity matrix.
The same is going on in your example above. You have $$4 \times 4$$ matrices in spinor space, $$3 \times 3$$ matrices in $$SU(3)$$ color space, and $$2 \times 2$$ matrices in isospin space. So really, everything is properly a $$24 \times 24$$ matrix. Every term should have a total of $$7$$ indices ($$1$$ Lorentz, $$2$$ spinor, $$2$$ color, $$2$$ isospin). Almost all of these indices are suppressed to keep the notation simple. (See here for more detail.)