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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

Description

• 

 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,Y10. Let spanZ1, Z2, ..., Zm1 be the orthogonal complement of Y1 ⅆetermined by Q. Write V = spanZ1, Z2,  .... , Zm1 Y1N.  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.

 

Examples

withDifferentialGeometry:withTensor:

 

Example 1.

Find the signature of 4 different quadratic forms defined on the tangent space at a point of a 4-dimensional manifold.

 

DGsetupx1,x2,x3,x4,M

frame name: M

(2.1)

 

First quadratic form.

g1evalDGdx1&tdx1+dx2&tdx2+dx3&tdx3+dx4&tdx4

g1:=dx1dx1+dx2dx2+dx3dx3+dx4dx4

(2.2)

QuadraticFormSignatureg1

D_x1,D_x2,D_x3,D_x4,,

(2.3)

We see that the quadratic form is positive-definite in all directions; it is a Riemannian metric.

 

Second quadratic form.

M > 

g2evalDGdx1&tdx1+dx2&tdx2+dx3&tdx3+dx4&tdx4

g2:=dx1dx1+dx2dx2+dx3dx3+dx4dx4

(2.4)
M > 

QuadraticFormSignatureg2

D_x2,D_x3,D_x4,D_x1,

(2.5)

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.

M > 

g3evalDG2dx1&sdx2+2dx3&sdx4

g3:=dx1dx2+dx2dx1+dx3dx4+dx4dx3

(2.6)
M > 

QuadraticFormSignatureg3

D_x1+D_x2,D_x3+D_x4,D_x1D_x2,D_x3D_x4,

(2.7)

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.

M > 

g4evalDGdx1&tdx1+dx2&tdx2dx3&tdx3

g4:=dx1dx1+dx2dx2dx3dx3

(2.8)
M > 

QuadraticFormSignatureg4

D_x1,D_x2,D_x3,D_x4

(2.9)

 

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.

M > 

QuadraticFormSignatureg4,output=dimensions

2,1,1

(2.10)

 

Example 2.

We calculate the signature of the quadratic forms restricted to some subspaces.

 

M > 

QuadraticFormSignatureg2,D_x1,D_x2,D_x3

D_x2,D_x3,D_x1,

(2.11)
M > 

QuadraticFormSignatureg4,D_x1,D_x3,D_x4

D_x1,D_x3,D_x4

(2.12)

 

Example 3.

Here we consider quadratic forms which depend upon a parameter.

 

M > 

DGsetupx1,x2,x3,x4,M

frame name: M

(2.13)

g5evalDGadx1&tensordx1dx2&tdx2+dx3&tdx3+dx4&tdx4

g5:=adx1dx1dx2dx2+dx3dx3+dx4dx4

(2.14)

Without further information on a it is not possible to compute the signature, that changes depending on the sign of a:

M > 

QuadraticFormSignatureg5

FAIL

(2.15)
M > 

QuadraticFormSignatureg5assuming0<a

D_x1&comma;D_x3&comma;D_x4&comma;D_x2&comma;

(2.16)
M > 

QuadraticFormSignatureg5assuminga<0

D_x3&comma;D_x4&comma;D_x1&comma;D_x2&comma;

(2.17)

 

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.

M > 

g6evalDGadx1&tensordx1+2dx1&sdx2dx2&tdx2+dx3&tdx3+dx4&tdx4

g6:=adx1dx1+dx1dx2+dx2dx1dx2dx2+dx3dx3+dx4dx4

(2.18)
M > 

QuadraticFormSignatureg6

FAIL

(2.19)

 

Now set the infolevel to 2.

M > 

infolevelQuadraticFormSignature2

infolevelDifferentialGeometry:-Tensor:-QuadraticFormSignature:=2

(2.20)
M > 

QuadraticFormSignatureg6

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

FAIL

(2.21)

 

We see that the signature depends on the sign of a&plus;1 

M > 

infolevelQuadraticFormSignature0

infolevelDifferentialGeometry:-Tensor:-QuadraticFormSignature:=0

(2.22)
M > 

QuadraticFormSignatureg6assuminga<1

D_x3&comma;D_x4&comma;D_x1&comma;D_x1aD_x2&comma;

(2.23)
M > 

QuadraticFormSignatureg6assuming1<a

D_x1+D_x2&comma;D_x3&comma;D_x4&comma;D_x2&comma;

(2.24)

See Also

DifferentialGeometry

Tensor

DGGramSchmidt

SubspaceType