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comment by SamEisenstat · 2023-07-25T16:21:42.183Z · LW(p) · GW(p)
We can then quotient by this relation to get a vector space
I think you're confusing two different parts here. There's a quotient of a vector space to get a vector space, which is done to embed $\mathcal{L}$ in a vector space. There's also something sort of like a projectivization, which does not produce a vector space. In the method I prefer, there isn't an explicit quotient, but instead just functions on the vector space that satisfy certain properties. (I could see being convinced to prefer the other version if it did improve the presentation.)
comment by SamEisenstat · 2023-07-25T16:10:34.197Z · LW(p) · GW(p)
of differences of lotteries
Is this supposed to be the square of the space of lotteries? The square would correspond to formal differences, but actual differences would be a different space.
The point of my construction with formal differences is that differences of lotteries are not defined a priori. If we embed $\mathcal{L}$ in a vector space then we have already done what my construction is for. This is all in https://link.springer.com/article/10.1007/BF02413910 in some form, and many other places.
Happy to talk more about this.
Replies from: SharkoRubio↑ comment by Jesse Richardson (SharkoRubio) · 2023-07-27T14:52:57.655Z · LW(p) · GW(p)
Yep I see what you mean, I've changed the setup back to what you wrote with V_1 and V_0. My main concern is the part where we quotient V_1 by an equivalence relation to get V, I found this not super intuitive to follow and I'd ideally love to have a simpler way to express it.
The main part I don't get right now: I see that 1/(c(v+ + w−))*(v+ + w−) and 1/(c(v+ + w−))*(v- + w+) are convex combinations of elements of L and are therefore in L, however it seems to me that these two things being the same corresponds to v+ + w- = v- + w+, which is equivalent to v + v- + w- = v- + w + w- which is equivalent to v=w. So doesn't that mean that v is equivalent to w iff v=w so our equivalence relation isn't doing anything? I'm sure I'm misunderstanding something here. Also just wanted to check, v and w should be in V_1 right, rather than V_0? Or is that the source of my confusion.
Hopefully I can try and find a way to express it once I fully understand it, but if you think what you wrote on Overleaf is the simplest way to do it then we'll use that.
Replies from: SharkoRubio, SamEisenstat↑ comment by Jesse Richardson (SharkoRubio) · 2023-07-27T17:33:21.749Z · LW(p) · GW(p)
I've tried to give (see on the post) a different description of an equivalence relation that I find intuitive and I think gives the space V as we want it, but it may not be fully correct.
Replies from: SamEisenstat↑ comment by SamEisenstat · 2023-07-28T03:09:00.730Z · LW(p) · GW(p)
I haven't read too closely, but it looks like the equivalence relation that you're talking about in the post sets elements that are scalar multiples of each other in equivalence. This isn't the point of my equivalence; the stuff I wrote is all in terms of vectors, not directions. My other top-level comment discusses this.
↑ comment by SamEisenstat · 2023-07-28T03:05:49.173Z · LW(p) · GW(p)
Yeah, this could be clearer. The point is that 1/(c(v+ + w−))*(v+ + w−) and 1/(c(v+ + w−))*(v- + w+) are formal sums of elements of L. These formal sums have positive coefficients which sum to 1, so they represent convex combinations. But their not equal as formal sums, only the results of applying the convex combination operation of L are equal.