The definition for the Poincare series of a graded algebra over a field $k$, $A=A_0 \oplus A_1 \oplus A_2 \oplus \cdot \cdot \cdot$ , is $P(A)= \sum_{i \geq 0} \text{dim}_k(A_i)t^i$. I'm trying to compute this formal sum for the case of $\text{gr}(SL(2))=k[a,b,c,d]/(ad-bc)$. I was hoping someone would check that I've done the computation correctly:
Let $B:= k[a,b,c,d]$ and $I:=(ad-bc)$, so that $\text{gr}(SL(2))=B/I$ with grading $(B/I)_i=B_i/(I \cap B_i)$. We can write $I= A_0 \cdot (ad-bc) \oplus A_1 \cdot (ad-bc) \oplus A_2 \cdot (ad-bc) \oplus \cdot \cdot \cdot$
From here we can write down the Poincare series, $$ \begin{align} P(A)&=\sum_{i \geq2} (\dim(A_i)-\dim(A_{i-2}))t^i \\ &= \sum_{i \geq 2}\dim(A_i)t^i-t^2\sum_{i \geq2}\dim(A_{i-2})t^{i-2} \\ &= -1-4t+\frac{1}{(1-t)^4}-\frac{t^2}{(1-t)^4}\\ &=\mathbf{\frac{1+t}{(1-t)^3}-(1+4t).} \end{align}$$
Does this look ok?
The Poincare series of your ring is easy to compute taking into account that you are moding out a polynomial ring by a homogeneous polynomial of degree $2$ (see here): it is $$\frac{1-t^2}{(1-t)^4}=\frac{1+t}{(1-t)^3}.$$