NICHOLAS MANTON in "Topological Solitons" says

"One may also regard the gauge potential as a connection on an $SU(2)$ bundle over $\mathbb{S}^4$, with field strength $F$. The fact that we can equally well regard the action as defined on $\mathbb{S}^4$ or $\mathbb{R}^4$ is because it is conformally invariant, so the field equation is the same in either case."

Why the the field equation is the same in $\mathbb{S}^4$ or $\mathbb{R}^4$?


Through stereographic projection $\mathbb R^n$ can be mapped to $S^n$ minus a point. If $\mathbb R^n$ is compactified by including a "point at infinity", the map is to the whole of $S^n$.

The following is for $\mathbb R^2$ and $S^2$ because they are easier to draw, but it generalizes to $\mathbb R^n$ and $S^n$ if you write out the transformations.

Put a 2-sphere in Euclidean 3-space with its "south pole" at the origin and let the tangent plane at this point be the $xy$-plane. For each point $P$ in the $xy$-plane, draw a line from $P$ to the north pole of the sphere (image from Wikimedia Commons):

Image of stereographic projection

The line will intersect the sphere in exactly one point, and each point on the sphere except the north pole corresponds to a point on the plane. Thus we get a map between $\mathbb R^2$ and $S^2$ minus a point. If we include a "point at infinity", it will be sent to the north pole.

One can show that this mapping is conformal, i.e., it preserves angles. Thus for any conformally invariant equation (i.e., it depends only on angles between vectors) we can view it equally well as an equation on $\mathbb R^n$ or on $S^n$.

The "point at infinity" can throw a wrench into this because one typically has some boundary conditions at infinity for the $\mathbb R^n$-theory and one needs to consider how to transfer these to the $S^n$ theory. This can depend on the topology of the solutions, but I'm sure that's what your book will teach you about.

  • $\begingroup$ I think i got. Correct me i am wrong. Since we want a finite action we can compactify $R^4$ into $S^4$, and since the action is conformally invariant we can take the same action but now in $S^4$ $\endgroup$ – amilton moreira May 1 '17 at 22:08

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