# The product of dg Lie algebras

I am trying to understand what are products and coproducts in the category of dg Lie algebras. I am okay with coproducts. For products, however, this Wikipedia article says that given $\mathfrak{g},\mathfrak{h}$ two dg Lie algebras, their product is given by the dg Lie algebra with underlying vector space given by the product $\mathfrak{g}\times \mathfrak{h}$ with bracket $$[(x,y),(x',y')] = ([x,y],[x',y'])$$ and differential $$d(x,y) = (dx,dy).$$ My problem is with this differential. It looks to me that it would have degree $-2$ instead of $-1$ as we would like. Am I missing something (such as using the correct grading on $\mathfrak{g}\times\mathfrak{h}$)?

• If you let $(x,y)$ have degree $|x|+|y|-1$, then $d(x,y)$ has degree $|x|-1+|y|-1=(|x|+|y|-1)-1$. I would guess this is the usual convention, for example when grading the fundamental groups with Whitehead bracket. – Pedro Tamaroff Apr 23 '16 at 13:23
• @PedroTamaroff I imagined it would be something like that. Thanks :) I'll check the details and verify it has the correct universal property. If you write your comment in an answer, I will accept it. Also, someone should add this detail to the Wiki article. – Daniel Robert-Nicoud Apr 23 '16 at 13:25
• @PedroTamaroff Ok, sorry it took me so much to give you a decent feedback - I was otherwise occupied. With your solution the differential has degree $-1$, but the bracket has degree $+1$, so we still have a big problem. Also, as I mentioned in the comment I have deleted, I don't see any natural way to define the projections from the product to the components... – Daniel Robert-Nicoud Apr 23 '16 at 16:18
• We're making it too complicated. The product of two graded vector spaces is their coproduct, so $(V\oplus W)_i = V_i\oplus W_i$. – Pedro Tamaroff Apr 23 '16 at 16:58
• @PedroTamaroff Ah, right, so then the pure elements of degree $k$ are sum of pure elements of degree $k$, and all makes sense. Thanks, that certainly works :) – Daniel Robert-Nicoud Apr 23 '16 at 20:47