Tensor[QuadraticFormSignature] - find the signature of a covariant, symmetric, rank 2 tensor
Calling Sequences
QuadraticFormSignature(Q, B, option)
Parameters
Q - a covariant, symmetric, rank 2 tensor, possibly degenerate
B - (optional) a list of vectors spanning a subspace of the vector space upon which the tensor Qis defined
option - keyword argument output = "dimensions"
Description
Examples
Let V be a vector space and let Q be a covariant, symmetric, rank 2 tensor (quadratic form) defined on V. The null space of Q is defined to be N = { X∈ V | QY,X = 0 for all Y ∈ V}. The vector space V decomposes into a direct sum
V = V+ ⊕V −⊕N , (*)
where Q is positive-definite on V+ and negative-definite on V− , that is, QX, X ≥ 0 for all X ∈ V+ and QX, X ≤0 for all X ∈ V−. Equality holds if and only if X = 0 . The subspace N is unique but the subspaces V+ and V− are not unique. Nevertheless, Sylvester's law of inertia states that the dimensions of V+ and V− are uniquely determined by Q.
The quadratic form Q is called non-degenerate if N = 0. In this case the dimensions of V+ and V− specify the signature of Q.
The command QuadraticFormSignature returns either a list of lists of vectors spanning the subspaces V+ ,V −, N or a list of their dimensions.
The algorithm that is used is as follows. Let B be a basis for V. First, calculate the null space N of Q . Use the command ComplementaryBasis to find vectors Xi ∈B such that
V = spanX1, X2, ... , Xm ⊕N.
Search through the set of vectors {Xi , Xi + Xj} to find a vector Y1 such that QY1,Y1≠0. Let spanZ1, Z2, ..., Zm−1 be the orthogonal complement of Y1 ⅆetermined by Q. Write V = spanZ1, Z2, .... , Zm−1 ⊕Y1⊕N. If QY1, Y1 >0 then Y1 ∈ V+ or Y1 ∈V−. Continue in this way to arrive at (*). Note that this method does not require the calculation of the eigenvalues/eigenvectors of Q.
If the quadratic form depends upon parameters, then the assuming facility may be useful in determining the sign of QY1, Y1 at each step in the algorithm.
with⁡DifferentialGeometry:with⁡Tensor:
Example 1.
Find the signature of 4 different quadratic forms defined on the tangent space at a point of a 4-dimensional manifold.
DGsetup⁡x1,x2,x3,x4,M
frame name: M
First quadratic form.
g1≔evalDG⁡dx1&tdx1+dx2&tdx2+dx3&tdx3+dx4&tdx4
g1:=dx1⁢dx1+dx2⁢dx2+dx3⁢dx3+dx4⁢dx4
QuadraticFormSignature⁡g1
D_x1,D_x2,D_x3,D_x4,,
We see that the quadratic form is positive-definite in all directions; it is a Riemannian metric.
Second quadratic form.
g2≔evalDG⁡−dx1&tdx1+dx2&tdx2+dx3&tdx3+dx4&tdx4
g2:=−dx1⁢dx1+dx2⁢dx2+dx3⁢dx3+dx4⁢dx4
QuadraticFormSignature⁡g2
D_x2,D_x3,D_x4,D_x1,
The quadratic form is positive-definite in the 3 directions [Dx2, Dx3, Dx4] and negative-definite in the 1 direction Dx1; it is a Lorentzian metric.
Third quadratic form.
g3≔evalDG⁡2⁢dx1&sdx2+2⁢dx3&sdx4
g3:=dx1⁢dx2+dx2⁢dx1+dx3⁢dx4+dx4⁢dx3
QuadraticFormSignature⁡g3
D_x1+D_x2,D_x3+D_x4,D_x1−D_x2,D_x3−D_x4,
The quadratic form is positive-definite in the 2 directions [Dx1 +Dx2 , Dx3 + Dx4] and negative-definite in the 2 directions [-Dx1 +Dx2 , Dx3 − Dx4] .
Fourth quadratic form.
g4≔evalDG⁡dx1&tdx1+dx2&tdx2−dx3&tdx3
g4:=dx1⁢dx1+dx2⁢dx2−dx3⁢dx3
QuadraticFormSignature⁡g4
D_x1,D_x2,D_x3,D_x4
The quadratic form is positive-definite in the 2 directions [Dx1 ,Dx2] and negative-definite in the direction [Dx3] and degenerate in the direction [Dx4]. Here are the dimensions of these spaces.
QuadraticFormSignature⁡g4,output=dimensions
2,1,1
Example 2.
We calculate the signature of the quadratic forms restricted to some subspaces.
QuadraticFormSignature⁡g2,D_x1,D_x2,D_x3
D_x2,D_x3,D_x1,
QuadraticFormSignature⁡g4,D_x1,D_x3,D_x4
D_x1,D_x3,D_x4
Example 3.
Here we consider quadratic forms which depend upon a parameter.
g5≔evalDG⁡a⁢dx1&tensordx1−dx2&tdx2+dx3&tdx3+dx4&tdx4
g5:=a⁢dx1⁢dx1−dx2⁢dx2+dx3⁢dx3+dx4⁢dx4
Without further information on a it is not possible to compute the signature, that changes depending on the sign of a:
QuadraticFormSignature⁡g5
FAIL
QuadraticFormSignature⁡g5assuming0<a
D_x1,D_x3,D_x4,D_x2,
QuadraticFormSignature⁡g5assuminga<0
D_x3,D_x4,D_x1,D_x2,
For more complicated examples, use infolevel to trace the testing performed by the QuadraticFormSignature procedure to see exactly at what point in the algorithm the procedure returns FAIL.
g6≔evalDG⁡a⁢dx1&tensordx1+2⁢dx1&sdx2−dx2&tdx2+dx3&tdx3+dx4&tdx4
g6:=a⁢dx1⁢dx1+dx1⁢dx2+dx2⁢dx1−dx2⁢dx2+dx3⁢dx3+dx4⁢dx4
QuadraticFormSignature⁡g6
Now set the infolevel to 2.
infolevelQuadraticFormSignature≔2
infolevelDifferentialGeometry:-Tensor:-QuadraticFormSignature:=2
The null space of the metric is
Testing vector: D_x1 The norm of this vector is: a Testing vector: D_x2 The norm of this vector is: -1 Testing vector: D_x1+D_x2 The norm of this vector is: a+1 Testing vector: D_x3 The norm of this vector is: 1 Testing vector: D_x1+D_x2 The norm of this vector is: a+1 Testing vector: D_x4 The norm of this vector is: 1 Testing vector: D_x1+D_x2 The norm of this vector is: a+1
We see that the signature depends on the sign of a+1
infolevelQuadraticFormSignature≔0
infolevelDifferentialGeometry:-Tensor:-QuadraticFormSignature:=0
QuadraticFormSignature⁡g6assuminga<−1
D_x3,D_x4,D_x1,D_x1−a⁢D_x2,
QuadraticFormSignature⁡g6assuming−1<a
D_x1+D_x2,D_x3,D_x4,D_x2,
See Also
DifferentialGeometry
Tensor
DGGramSchmidt
SubspaceType
Download Help Document