We now turn to understanding the kernel and image of the vector Laplacian. We examine two cases: closed manifolds and compact manifolds with boundary.
3.8.1
M closed
ConsiderM3, a closed 3-manifold. DecomposeV F(M) =F K(M)
⊕HK(M)⊕G(M) by the Hodge Theorem. How does the vector Laplacian act on each subspace?
Corollary3.23 implies that the kernel of the vector Laplacian is precisely the harmonic knotsHK. For a simply-connected manifold, such as the three-sphere, the vector Laplacian has a trivial kernel. Consider a fluxless knotV. ThenV is divergence free and so isL(V) =−∇ × ∇ ×V. The curl operator maps the space of fluxless knots bijectively onto itself, hence so does the vector Laplacian. In Proposition 3.21 we showed that the vector Laplacian sends gradients to gradients, but is it onto? Consider a gradient, ∇f; we seek another gradient ∇u such that L(∇u) = ∇f. Note L(∇u) =∇(∆u). On a closed manifold, the scalar Laplacian is invertible for functions with average value 0; i.e., there exists a functionusuch that ∆u=f−[f]. Then∇(∆u) =∇f, and we conclude thatL(G) =G.
In the paragraphs above, we have proven the following theorem:
Theorem 3.24. LetM be closed. The vector Laplacian respects the Hodge decomposition ofV F(M). Its kernel is HK(M), and it maps the other subspaces bijectively to themselves,
L(F K) = F K L(HK) = 0
For a closed manifold with H1(M,R) = 0, such as the three-sphere, the vector Laplacian has
trivial kernel and mapsV F(M) bijectively to itself. We use that fact to define an inverse to L in section 3.9.
3.8.2
M compact with boundary
LetM3be compact with piecewise smooth∂M. We assume that the Hodge Decomposition Theorem
for vector fields on M is analogous to Theorem 3.5, i.e., the space V F(M) decomposes into five orthogonal subspaces: V F(M) =F K⊕HK⊕CG⊕HG⊕GG.
Theorem 3.25. The vector Laplacian onV F(M)has the following kernel: kerL = HK⊕CG⊕HG⊕CL ,
whereCL⊂F K⊕GG is defined to be the space
CL ={Vf− ∇gf |Vf ∈F K,∇gf ∈GG,−∇ × ∇ ×V =∇(∆gf)∈CG} .
Proof. Any vector field V that lies in the kernels of both the curl and divergence operators will necessarily haveL(V) = 0. So the kernel ofLincludes the subspaceHK⊕CG⊕HG.
Now consider the space of fluxless knots. ForV ∈F K, then we haveL(V) =−∇ × ∇ ×V. A subspace ofF K is sent by the curl operator into the spaceHK⊕CG; call this subspaceF KL⊂F K.
For V ∈F KL, its curl∇ ×V lies in the kernel of curl, so L(V) = 0. Thus, F KL must lie in the
kernel ofL.
Furthermore, these are the only fluxless knots in the kernel. All other fluxless knotsV have some component of their curl that lies inF K; thus∇ × ∇ ×V =L(V) does not vanish.
Let’s now examine ∇g ∈ GG; we see that L(∇g) = ∇∆g. For this to vanish, ∆g must be a constant. So the subset of GG in the kernel is {∇g ∈ GG(M) | ∆g = constant}; denote this as GGL.
the grounded gradients onto the curly gradients. Thus any curly gradient∇f has two corresponding preimages via the vector Laplacian. We defineCL⊂F K⊕GGas the subspace of differences of these
preimages: any vector field in CL is the difference of a fluxless knot Vf and a grounded gradient
∇gf, which are both mapped to the same curly gradient via the vector Laplacian, i.e.,
L(Vf) =L(∇gf) =∇f ∈CG .
As a subset ofF K⊕GG, the spaceCLis orthogonal to the remainder of the kernel,HK⊕CG⊕
GG. Moreover, the subspacesF KL andGGL are trivially contained inCL.
Gathering the pieces, we have that kerL=HK⊕CG⊕HG⊕CL; the theorem is proven.
Theorem 3.26. ForM as above, the vector Laplacian is surjective, i.e.,Image (L) =V F(M). Proof. We first showL is onto the space of fluid knots,K(M). For anyW ∈Image (curl), we can find a X ∈ F K such that∇ ×X =W; we can also find V ∈ F K such that−∇ ×V =X. Thus L(V) =−∇ × ∇ ×V =W. In conclusion,L(F K) =F K⊕HK⊕CG.
LetW =∇gbe an arbitrary gradient onM, wheregcan be chosen so that its average value on M is zero. We seek a gradient ∇u∈GG such thatL(∇u) =∇g. Since ∇u ∈GG, we have that u|∂M = 0. SinceL(∇u) =∇∆u, the desired condition is tantamount to the Dirichlet problem:
∆u = g (onM) u|∂M = 0
The Dirichlet problem has a unique solution, and we conclude L(GG) =CG⊕HG⊕GG. Having calculated the image of the subspacesF K andGG, we recall that the other three subspacesHK⊕ CG⊕HGlie in the kernel ofL, as proved in Theorem 3.25. Therefore, the result is shown:
L(V F(M)) =L(F K)∪L(GG)
=F K⊕HK⊕CG ∪ CG⊕HG⊕GG =V F(M)