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0 Let $Y$ be a compact Kähler manifold of dimension $n$ and an action of $S_3$ on $Y$ is a homomorphism from $S_3$ to the group Homeo $(Y)$ of all homeomorphisms from $Y$ to itself. Now there is an $S_3$ action on Alb( $Y$ ):= $H^{n-1,n}(X)/im(H^{2n-1}(X,mathbb{Z}))$ , induced by the action on $Y$ . Let $X$ be the fixed locus of Alb( $Y$ ) by this $S_3$ action. Then why is $X$ a subtorus? algebraic-geometry algebraic-topology kahler-manifolds share | cite | improve this question edited Jan 5 at 1:23 6666 asked Jan 4 a

8 2 Let’s say we have a current wire with a current $I$ flowing. We know there is a field of $B=frac{mu_0I}{2pi r}$ by using Ampère's law, and a simple integration path which goes circularly around the wire. Now if we take the path of integration as so the surface spans doesn’t intercept the wire we trivially get a $B=0$ which is obviously incorrect. I see that I have essentially treated it as if there is no current even present. But a similar argument is used in other situations without fault. Take for example a conducting cylinder with a hollow, cylindrical shaped space inside. By the same argument there is no field inside. To further illustrate my point, the derivation of the B field inside of a solenoid requires you to intercept the currents. You can’t simply do the loop inside of the air gap. This, at