# The shift-invariance operator

I am reading the paper "Algebraic Signal Processing Theory: Foundation and 1-D Time". In page one, the first paragraph of second column, I did not get why shift invariance forces the algebra $$A$$ to be commutative? Is it taking about a particular signal model $$(A,M,\Phi)$$ or it is considered a general case?

The central notion is that the considered algebras are generated by a shift operator $$\mathcal{S}$$. Every element (filter) of the algebra should be (by definition) generated by a (finite) number of algebraic operations (addition, product, scalar multiplication) involving the generator and its inverse. This is explained in more details in Section II-A, around Equation (30), and footnote 19. So if the shift commutes with any filter, since any filter is made of a combination of shifts and inverse shifts, then any filter commutes with any filter.
For instance, with standard discrete filters this unit element would be the Kronecker $$\delta_0$$ (the "unit" algebra element relative to the multiplicative $$\cdot$$ operation of the algebra, here the convolution). A (finite support) filter would be written as sums and multiplies of shifts, with scalars corresponding to filter taps: $$h_0S^0(\delta_0)+h_1S(\delta_0)+h_2S^2(\delta_0)+\cdots$$ and commute with basic shifts.
• Thanks for your attention. Actually, I am positive with your idea. Just some points need to be noticed. 1) Every singly generated algebra is commutative. So, I do not get why it is emphasized in the paper that, when the operator is shift-invariance then the algebra $A$ is commutative. 2) In abstract case, in an algebra, what is meaning of shift operator? shift-invariance operator?!! – Ali Bagheri Dec 14 '20 at 6:32