Let A be the set of all binary time series. (for example, all spike trains between time t_1 and t_2) I am looking for an operation *, such that (A,*) form a group.
Is the set of all spike trains form an abelian group?
That depends at least in part on your definition of *. I see no obvious choice, so we'll have to specify one. It will also make our task easier if we look only at sets of the same length.
Let B be the set of all binary time series of length L. For clarity, an example element from B for L = 3 could be b1 = (b11,b12,b13) = (T,F,T). Let * be the bitwise AND operator, i.e. T*T=T, T*F=F and F*F=F and define b1*b2 = (b11*b21, b12*b22, b13*b23). The set B is closed under * and contains the identity under *, I = (T,T,T), and the * operator is associative. However, there exists an element (F,F,F) such that there exists no inverse element. So, (B,*) is not a group under this definition.
Similarly, if we define * to be the bitwise OR, we fail to find an inverse for (T,T,T).
However, if we let * be the bitwise XOR operator(T*T = F, T*F = T, F*F=F, then (B,*) is a group. Here the identity element is (F,F,F), and (B,*) is associative, closed under * and closed under inverse. (B,*) is also abelian for this definition of *.
Yes. XOR operator works. But first, we need to have a norm (length) on the set of all binary time series. So before we see it as a group, we should see it as a metric space. Is there any way to remove the length condition?
Thank you.
It is not clear to me why you require a metric space. You can have a group defined over a set without a metric.
If for some other reason, you require a metric space, then yes you would need a metric in order to have a metric space.
Let me explain. you wrote: "Let B be the set of all binary time series of length L". So, First we need a length to make a group of time series. Is it true?
Is it possible to define operation * on B such that L is unknown for us?
With (B,*) as XOR was defined, all members of the group would have the same length, but we don't need to know what the length is. Under the current definition, (T,F) * (T,T,F) has no obvious meaning.
It may be possible to extend the definition of the group to contain elements with different lengths. If one defines C to be the set of all (c1,c2,..,cn,..), with ci in {1,0} and cj = 0 for all j > N for some finite N. Then (C,*) is an abelian group.
The requirement that cj = 0 for all j > N for some finite N is not necessary for a group, but is I think necessary to define a metric.
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