Computations for the paper: Sufficient conditions for the existence of limiting Carleman weights (section 7)¶

Pablo Angulo-Ardoy, Daniel Faraco, Luis Guijarro

This notebook only covers the example in section 7

M = Manifold(4, 'M', r'\mathcal{M}', start_index=1)
print M
X.<t,x,y,z> = M.chart()
print X
g = M.riemannian_metric('g')
print g

4-dimensional differentiable manifold M
Chart (M, (t, x, y, z))
Riemannian metric g on the 4-dimensional differentiable manifold M

#You can use two generic functions
#F(s) = function('F', s)
#H(s) = function('H', s)
#g[1,1], g[2,2], g[3,3], g[4,4] = 1, 1, F(x), H(x)
#g.display()

assume(x>0)
g[1,1], g[2,2], g[3,3], g[4,4] = 1, 1, x^3, 1/x
g.display()

g = dt*dt + dx*dx + x^3 dy*dy + 1/x dz*dz

show(g[:])

The Weyl tensor has type C¶

%time W = g.weyl()

CPU times: user 12 s, sys: 340 ms, total: 12.4 s
Wall time: 14.5 s

W4 = W.down(g,0)

Frame = X.frame()
def g_norm(v):
    return sqrt(g(v,v))
vs  = [Frame[i] for i in [1..4]]
ovs = [v/g_norm(v) for v in vs]
v1,v2,v3,v4 = ovs

N = 4
INDEX_PAIRS = [tuple(c) for c in Combinations([1..N], 2)]

#Bivectors
BASIS = [(vi*vj).antisymmetrize(0,1) for vi,vj in Combinations(ovs, 2)]
for elem in BASIS:
    print elem.display()
d = 6
v12 = BASIS[0]
v13 = BASIS[1]

def WO(bv1, bv2):
    r = 0
    for i,j in INDEX_PAIRS:
        c1 = bv1[i,j]
        for k,l in INDEX_PAIRS:
            c2 = bv2[k,l]
            r += c1*c2*W4[i,j,k,l]
    return r

1/2 d/dt*d/dx - 1/2 d/dx*d/dt
1/2/x^(3/2) d/dt*d/dy - 1/2/x^(3/2) d/dy*d/dt
1/2*sqrt(x) d/dt*d/dz - 1/2*sqrt(x) d/dz*d/dt
1/2/x^(3/2) d/dx*d/dy - 1/2/x^(3/2) d/dy*d/dx
1/2*sqrt(x) d/dx*d/dz - 1/2*sqrt(x) d/dz*d/dx
1/2/x d/dy*d/dz - 1/2/x d/dz*d/dy

MM = matrix(SR, d)
for i, bvi in enumerate(BASIS):
    for j,bvj in enumerate(BASIS):
        MM[i,j] = WO(bvi, bvj).expr()

We notice that the Weyl tensor is of type C

show(MM)

Is this a product of surfaces?¶

On account of theorem 1.7, the manifold can only split as a product of surfaces if the factors are tangent to $\langle\partial_t, \partial_x\rangle$ and $\langle\partial_y, \partial_z\rangle$ .

We use theorem 2.4 to check if these are conformal factors

n = g.connection()

show(n.display())

We want to check whether the surfaces $<\partial_t, \partial_x>$ are $<\partial_y, \partial_z>$ are umbilic

#An orthonormal basis
v1,v2,v3,v4 = ovs

#All this functions must be identically 0 (and they are)
print g(n(v1)(v2),v3).display()
print g(n(v1)(v2),v4).display()
print g(n(v3)(v4),v1).display()
print g(n(v3)(v4),v2).display()

M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0

#These two functions must agree on every point (and they do)
print g(n(v1)(v1),v3).display()
print g(n(v2)(v2),v3).display()

M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0

#These two functions must agree on every point (and they do)
print g(n(v1)(v1),v4).display()
print g(n(v2)(v2),v4).display()

M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0

#These two functions must agree on every point (and they do)
print g(n(v3)(v3),v1).display()
print g(n(v4)(v4),v1).display()

M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0

#These two functions must agree on every point (but THEY DON'T!!)
print g(n(v3)(v3),v2).display()
print g(n(v4)(v4),v2).display()

M --> R
(t, x, y, z) |--> -3/2/x
M --> R
(t, x, y, z) |--> 1/2/x

Thus the manifold is not conformal to a product of surfaces.

We check if the coordinate vector fields are LCW¶

The vector fields $\partial_t$ , $\partial_y$ and $\partial_z$ are obviously LCWs. We consider $\partial_x$ :

#A sanity check: second fundamental form of partial_t^\perp
II = matrix(SR, 3)
for j,v_j in enumerate((v2,v3,v4)):
    for k,v_k in enumerate((v2,v3,v4)):
        II[j,k] = g(n(v_j)(v_k),v1).expr()
print II

[0 0 0]
[0 0 0]
[0 0 0]

However, the second fundamental form of $\partial_x^\perp$ is not a multiple of the identity

#Second fundamental form of partial_x^\perp
II = matrix(SR, 3)
for j,v_j in enumerate((v1,v3,v4)):
    for k,v_k in enumerate((v1,v3,v4)):
        II[j,k] = g(n(v_j)(v_k),v2).expr()
show(II)

We find all LCWs¶

The analysis of the Weyl tensor shows that any possible conformal factor must be contained in either $\langle\partial_t, \partial_x\rangle$ or $\langle\partial_y, \partial_z\rangle$ .

We first seek conformal factors in $\langle\partial_t, \partial_x\rangle$ :

print g(n(v3)(v3),v1).display()
print g(n(v3)(v4),v1).display()
print g(n(v4)(v4),v1).display()

M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 0

print g(n(v3)(v3),v2).display()
print g(n(v3)(v4),v2).display()
print g(n(v4)(v4),v2).display()

M --> R
(t, x, y, z) |--> -3/2/x
M --> R
(t, x, y, z) |--> 0
M --> R
(t, x, y, z) |--> 1/2/x

The function $Z\rightarrow g(\nabla_Z Z,v_1)$ is identically zero, but $Z\rightarrow g(\nabla_Z Z,v_2)$ is not a multiple of the identity.

Thus the only possible LCW is $(x,y,z,t)\rightarrow t$ , which obviously is

We now look for conformal factors in $\langle\partial_y, \partial_z\rangle$ ¶

We compute the second fundamental form of $O^\perp$ ¶

We define $O = cos(\alpha)\: v_3+sin(\alpha)\:v_4$ and $P = -sin(\alpha)\:v_3+cos(\alpha)\:v_4$ for a generic function $\alpha(x,y,z,t)$ . We want to find equations for $\alpha$ so that $O^\perp$ is umbilic:

alpha = function('alpha', t, x, y, z)
O = cos(alpha)*v3+sin(alpha)*v4
P = -sin(alpha)*v3+cos(alpha)*v4

#Segunda forma fundamental de O^\perp
II = matrix(SR, 3)
for j,v_j in enumerate((v1,v2,P)):
    for k,v_k in enumerate((v1,v2,P)):
        II[j,k] = g(n(v_j)(v_k),O).expr()

/projects/sage/sage-6.10/local/lib/python2.7/site-packages/IPython/core/interactiveshell.py:3066: DeprecationWarning: Calling function('f',x) is deprecated. Use function('f')(x) instead.
See http://trac.sagemath.org/17447 for details.
  exec(code_obj, self.user_global_ns, self.user_ns)

for j,v_j in enumerate((v1,v2,P)):
    for k,v_k in enumerate((v1,v2,P)):
        if II[j,k]:
            print 'II_{%d,%d}=%s'%(j, k, II[j,k].simplify_full())

II_{1,2}=-2*cos(alpha(t, x, y, z))*sin(alpha(t, x, y, z))/x
II_{2,0}=-D[0](alpha)(t, x, y, z)
II_{2,1}=-D[1](alpha)(t, x, y, z)
II_{2,2}=-(x^2*cos(alpha(t, x, y, z))*D[3](alpha)(t, x, y, z) - sin(alpha(t, x, y, z))*D[2](alpha)(t, x, y, z))/x^(3/2)

Remark: This is not symmetric! This is because for some choices of $\alpha$ the distribution $Z^\perp$ is not integrable. The distribution is umbilic if and only if II is a multiple of the identity, so an umbilic distribution is integrable.

The entry $II_{1,2}=-\frac{2}{x}\: cos(\alpha(t, x, y, z))\: sin(\alpha(t, x, y, z))$ is very explicit: $\alpha$ must be a multiple of $\pi/2$ at every point. Since $\alpha$ is a continuous function, P must be either v1 or v2.