Affine Spaces are..?

Fredrik

No. I said that for each s in S, we define ##\phi_s:V\to S## by ##\phi_s(v)=\phi(s,v)## for all v in V. So each ##\phi_s## takes a member of V to a member of S.

Suppose that X,Y,Z are vector spaces. A map ##f:X\times Y\to Z## is said to be bilinear if for all ##(x,y)\in X\times Y## both of the maps ##f_x:Y\to Z## and ##f_y:X\to Z## defined by ##f_x(v)=f(x,v)## for all v in Y and ##f_y(u)=f(u,y)## for all u in X, are linear.

When we define ##\phi##, we can't say that it's bilinear, since S isn't even a vector space. However, once we have used one of the ##\phi_s## maps to turn S into a vector space, then it makes sense to ask if ##\phi## is bilinear with respect to that vector space structure on S. It should be easy to check if it is, but I haven't done it myself, so I don't know for sure.


It doesn't work, unless we define those sums. We could do this using the function ##\phi## (just define ##v+s=\phi(v,s)##), but this is kind of pointless. We're not interested in those sums. We're only interested in differences s'-s (where s and s' are both in S) and we define those by saying that s'-s is the unique v such that ##\phi(v,s)=s'##.

Tenshou

Oh okay, thank for cleaning this up for me.