For Example 2 / Prop 35, would this model also work?
Define W to be the factor corresponding to the question “are the second and third bits equal or not?” Then ((Ω,{X,V,W}),id) is a model of Ω. I believe this is consistent with D:
For O={(X,V,{Ω}),(X,Z,Y),(V,Z,Y)}:
We have hF(X)={X} and hF(V)={V} for the first condition.
We have hF(X∣Y)={X} and hF(Z∣Y)={W} for the second condition.
We have hF(V∣Y)={V} and hF(Z∣Y)={W} for the third condition.
ForN={(X,Z,{Ω}),(V,Z,{Ω}),(Z,Z,Y)}.
We have hF(Z)={X,V,W} for the first and second conditions.
Yeah, also note that the history of X given Y is not actually a well defined concept. There is only the history of X given y for y∈Y. You could define it to be the union of all of those, but that would not actually be used in the definition of orthogonality. In this case hF(X|y), hF(V|y), and hF(Z|y) are all independent of choice of y∈Y, but in general, you should be careful about that.
Yeah, fair point. (I did get this right in the summary; turns out if you try to explain things from first principles it becomes blindingly obvious what you should and shouldn’t be doing.)
For Example 2 / Prop 35, would this model also work?
Define W to be the factor corresponding to the question “are the second and third bits equal or not?” Then ((Ω,{X,V,W}),id) is a model of Ω. I believe this is consistent with D:
For O={(X,V,{Ω}),(X,Z,Y),(V,Z,Y)}:
We have hF(X)={X} and hF(V)={V} for the first condition.
We have hF(X∣Y)={X} and hF(Z∣Y)={W} for the second condition.
We have hF(V∣Y)={V} and hF(Z∣Y)={W} for the third condition.
For N={(X,Z,{Ω}),(V,Z,{Ω}),(Z,Z,Y)}.
We have hF(Z)={X,V,W} for the first and second conditions.
We have hF(Z∣Y)={W} for the third condition.
I think that works, I didn’t look very hard. Yore histories of X given Y and V given Y are wrong, but it doesn’t change the conclusion.
Yeah, both of those should be {X,V}, if I’m not mistaken (a second time).
Yeah, also note that the history of X given Y is not actually a well defined concept. There is only the history of X given y for y∈Y. You could define it to be the union of all of those, but that would not actually be used in the definition of orthogonality. In this case hF(X|y), hF(V|y), and hF(Z|y) are all independent of choice of y∈Y, but in general, you should be careful about that.
Yeah, fair point. (I did get this right in the summary; turns out if you try to explain things from first principles it becomes blindingly obvious what you should and shouldn’t be doing.)