Fundamental Symmetries in Minkowski Krein Space

Posted on 2026/05/20 by arkajad

It is always good to have at least one example. I am reading the Preface to A Course in Functional Analysis by John B. Conway (Springer, 1985), where the author notes:

Historically, mathematics has gone from the particular to the general—not the reverse. There are many reasons for this, but certainly one reason is that the human mind resists abstractions unless it first sees the need to abstract.

The concept of a Krein space is an abstraction, and in this series I am developing this abstraction. Why do we need it? For what purpose? I have several examples in mind. One is the Minkowski space of special relativity, which is a real Krein space. In my previous posts we encountered Clifford algebras and complex structures, in particular complex spaces with Hermitian scalar products of signatures (1,1) and (2,2), which arise naturally when studying conformal symmetries.

In this post we will take a closer look at the Minkowski space introduced in Example 1 and Example 2 in Krein spaces – first steps. I will assume that the reader has done the exercises following that example.

Minkowski space.

Let X=\mathbb{R}^{3,1} be standard Minkowski spacetime with coordinates x^1,x^2,x^3,x^4=ct. Let with G be its metric tensor

    \[ G=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-1\end{pmatrix}\]

and the Minkowski scalar product

    \[\langle x,y\rangle =x^TGy=x^1y^1+x^2y^2+x^3y^3-x^4y^4.\]

Then J=G is a fundamental symmetry: we define

    \[(x,y)_J=\langle x,Jy\rangle =x^TGJy=x^Ty,\]

and since J=G, and G^2=I. we obtain

    \[ (x,y)_J=x^1y^1+x^2y^2+x^3y^3+x^4y^4.\]

This is the usual construction turning the indefinite Minkowski form into a positive-definite inner product via a fundamental symmetry. See e.g.  “Krein-Space Framework: Theory & Applications“.

Let L be any Lorentz matrix, that is a real 4\times 4 matrix satisfyng

    \[L^TGL=G.\]

If we define J'=LJL^{-1}, then J'=LL^TG, and J' is again a fundamental symmetry. Thus Lorentz transformations (in particular Lorentz boosts)  generate, in general, new fundamental symmetries.

We now make next step and specify L. First notice that for pure spatial rotation the matrix L is orthogonal, L^TL=I. Therefore for pure rotations we do not obtain a new fundamental symmetry J', because in that case J'=J.  To obtain genuinely new fundamental symmetries we must include Lorentz boosts—transformations to reference frames moving with nonzero velocity with respect to the original one. (cf. Lorentz Transformations in Special Relativity).

Let us think about the physical significance of this observation. What does a fundamental symmetry of a Krein space do? It replaces an indefinite metric by a positive-definite one: in our particular case it replaces Minkowski geometry by a Euclidean geometry. This is equivalent to making the time coordinate imaginary instead of real. But the rate of time depends on the speed of the observer—hence the “twin paradox” and related effects. The splitting of spacetime into “space” and “time” is different for observers in relative motion. Fundamental symmetries precisely encode such splittings; they pick out different space–time decompositions compatible with the same underlying indefinite metric.

In quantum theory, replacing real time by imaginary time—replacing spacetime geometry by Euclidean geometry—is usually called “Wick rotation”. Stephen Hawking used this technique extensively in his work on cosmology. It is sometimes very helpful, but one must be careful, because the resulting Euclidean geometry depends on the observer’s reference frame. Peter Woit has devoted several posts to this subject on his most interesting blog Not Even Wrong:

A pure boost

Let us choose L to be a pure boost in the z– direction:

    \[ L=L(\zeta)=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&\cosh\,\zeta&\sinh\,\zeta\\ 0&0&\sinh\,\zeta&\cosh\,\zeta\end{pmatrix} \]

Note that \zeta=\tanh^{-1}\beta,\,\beta=v/c. The parameter \zeta is sometimes called the rapidity.

Let us compute J'=LL^TG. In our case L is symmetric, L^T=L, so L^TL=L^2. The matrices L(\zeta) form a one-parameter group,

    \[ L(\zera_1)L(\zeta_2)=L(\zeta_1+\zeta_2),\]

as can be verified by explicit calculations with \sinh and \cosh. Therefore

    \[ L^TL=l(2\zeta)=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&\cosh\,2\zeta&\sinh\,2\zeta\\ 0&0&\sinh\,2\zeta&\cosh\,2\zeta\end{pmatrix} .\]

This J'=LL^TG is given by

    \[J'=\begin{pmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&\cosh\,2\zeta&-\sinh\,2\zeta\\ 0&0&\sinh\,2\zeta&-\cosh\,2\zeta\end{pmatrix} .\]

In Krein spaces – a quantum-theoretical monad method, Proposition 2 sates that JJ' and J'J are positive self-adjoint operators on both X_J and X_{J'}. Let us see how this works in our particular case.

The space X_J is \mathbb{R}^4 endowed with the standard Euclidean scalar product, as we have seen it above. What is X_{J'} in our case? The scalar product in X_{J'} is defined by

    \[ (x,y)_{J'}=\langle x,J'y\rangle =x^TGL^TLGy.\]

Using J′=LL^TG and the Lorentz property L^T GL=G, we can rewrite this as

    \[(x,y)_{J'}=x^T GLL^TGy=(LGx)^T LGy.\]

This shows directly that (\cdot,\cdot)_{J′}​ is positive definite: setting z=L^T y. We then have:

    \[ (x,x)_{J'}=z^T z\geq 0,\]

with equality only for z=0, hence only for x=0. This positivity property is not so obvious  if we look  at the explicit formula, which can be computed  as

    \begin{align*} &(x,x)_{J'}=\\&(x^1)^2+(x^2)^2+((x^3)^2+(x^4)^2)\cosh \,2\zeta-2x^3x^4\sinh\,2 \zeta,\end{align*}

containing apparently “dangerous” mixed term that  can be  negative.

The operator T=JJ'

Let us denote

    \[ T= JJ'.\]

Since J^2=J'^2=I, we have

    \[ J'J = T^{-1}.\]

(Indeed,T=JJ' implies T^{-1}=J'^{-1}J^{-1}=J'J..)

Spectral decomposition

We know that T and  T^{-1} are self-adjoint positive operators on both X_J and X_{J'}. Let us analyze their spectral decomposition.

Linear algebra textbooks typically discuss the spectral theorem only in complex spaces, because we want to have n solutions for the eigenvalue equation of an n\times n matrix, and this property of polynomials is guaranteed  over \mathbb{C}, but may fail over \mathbb{R}. A notable exception is  S. Axler, “Linear Algebra Done Right“, where, on p. 136, 2nd edition, we find

Real Spectral Theorem: Suppose that V is a real inner-product space and T\in\mathcal{L}(V). Then V has an orthonormal basis consisting of eigenvectors of T if and only if T is selfadjoint.

Our T is self-adjoint (indeed, positive), so we should be able to find four eigenvalues and four eigenvectors. Moreover, these eigenvectors are orthogonal both in X_J and in X_J'.

Finding the eigenvectors of T is a simple exercise. Here is one convenient choice:

    \[ v_1=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}, v_2=\begin{pmatrix}0\\1\\0\\0\end{pmatrix}, v_3=\begin{pmatrix}0\\0\\1\\1\end{pmatrix}, v_4=\begin{pmatrix}0\\0\\-1\\1\end{pmatrix}.\]

Exercise 1. Verify the above statement.

Their norms in X_J are 1,1,\sqrt{2},\sqrt{2}. To find their norms in X_{J'}, a short computation is needed. The result is:

    \[ \Vert v_1\Vert_{J'}=\Vert v_2\Vert_{J'}=1.\]

    \[ \Vert v_3\Vert_{J'}=\sqrt{2}\,\exp(-\zeta),\, \Vert v_4\Vert_{J'}=\sqrt{2}\,\exp(\zeta).\]

Light cone surprise

Eigenvectors v_3 and v_4, corresponding to eigenvalues \neq 1, are on the light cone of Minkowski space: they are isotropic vectors for the Minkowski scalar product \langle \cdot,\cdot\rangle .  When I discovered this fact while writing this post, I was taken by surprise and asked myself: “Why?”. Is it a general feature valid in any Krein space, or is it just a coincidence specific to our example?

It is not difficult to find a simple argument suggesting that it must be so in general. However, because the argument is so simple, I am still suspicious about its validity. Here it is.

Proposition 1 Eigenvectors of JJ' corresponding to eigenvalues \lambda\neq 1 are always isotropic vectors in (X,\langle \cdot,\cdot\rangle ).

Proof. Suppose JJ'v=\lambda v with \lambda\neq 0. Then, since J'J=(JJ')^{-1}  J'Jv=(1/\lambda) v, we have

    \[ \langle v,v\rangle =\langle JJ'v,J'Jv\rangle .\]

Now we use the fact that J and J' are Hermitian with respect to the indefinite scalar product:

    \[\langle Jx,y\rangle =\langle x,Jy\rangle ,\quad \langle J'x,y\rangle =\langle x,J'y\rangle .\]

Thus

    \begin{align*} \langle v,v\rangle &=\langle JJ'v,J'Jv\rangle =\langle J'JJ'v,Jv\rangle =\langle JJ'JJ'v,v\rangle \\&=\lambda^2\langle v,v\rangle .\end{align*}

Since, by assumption, \lambda^2\neq 1, it follows that \langle v,v\rangle =0.

So, at least formally, eigenvectors associated with eigenvalues \lambda\neq 1 are isotropic.

Exercise 2. Is the above “proof” valid for a general Krein space? (I am not sure about the answer).

What next?

Next interesting, finite-dimensional, case of interest is \mathbb{C}^{2,2} – the space of conformal twistors.

However, my aim is to develop a general theory that also covers the case of infinitely many dimensions, both real and complex. In infinite dimensions new complications arise. The spectrum of a bounded self-adjoint operator may be purely continuous, with no eigenvectors of finite norm. Topology (weak, strong, norm) starts to play a role in an essential way. I also want to cover non-separable Hilbert spaces, just in case.

We then need to use full spectral theory, which is usually developed in detail only in the complex case; the real case is often treated more briefly. To bridge this gap we will need to introduce complexification. So there is still some work to be done. In the next post we will discuss the complexification machinery.

After note (20-05-26 13:50)

I think I have found a better, more elegant proof of Proposition 1. The new proof should be adaptable to the continuous spectrum case in infinite dimensions. For a finite dimension it goes as follows: first we prove a simple lemma:
Lemma. JTJ = T^{-1}.
Proof. JTJ=JJJ’J=J’J=T^{-1}.

Then comes corollary:

Corollary. If v is an eigenvector to the eigenvalue \lambda, the Jv is an eigenvector to the eigenvalue 1/\lambda.
Proof. Follows immediately from the lemma.

Now we prove Proposition 1: Assume v is an eigenvector of T to an eigenvalue \lambda\neq 1. We have

    \[ \langle v,v\rangle =(v,Jv)_J=0\]

since vectors v and Jv are eigenvectors belonging to two different eigenvalues, and, by the Spectral Theorem,  are orthogonal in X_J.

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15 Responses to Fundamental Symmetries in Minkowski Krein Space

  1. Bjab says:
    2026/05/21 at 09:44

    Wzór pod “G be its metric tensor” ->
    zbyt dużo wymiarów

    Wzór pod “matrix satisfyng” ->
    +?

    Note that \zeta=\tanh^{-1}\beta,\,\beta=v/ct he parameter \zeta is sometimes ->
    spacje

    The matrices L(\zeta form a one-parameter ->
    nawias

  2. Bjab says:
    2026/05/21 at 11:19

    Dziwne napisy pojawiają się pod “and the Lorentz property”

    a także
    ziwne napisy pojawiają się pod “This shows directly that”

  3. Bjab says:
    2026/05/21 at 12:32

    Jakiś nawiasik przed “Let us see how”

    self-adjoint positive operators on both X^J ->
    X_J

    these eigenvectors are orthogonal both in X_J and in X_J’. ->
    prim ?

    Powyżej “Light cone surprise” jakieś e jakieś zeta?

  4. Bjab says:
    2026/05/21 at 12:36

    eigenvalues \neq 1 1, ->
    raczej 1 a nie 11

    specific yo our ->
    specific to our

    Proof. Suppose JJ’v=\lambda v with \lambda\neq 0. Then, since J’J=(JJ’)^{-1} J’Jv=(1/\lambda v, we have ->
    brak nawiasu zamykającego

  5. Bjab says:
    2026/05/21 at 12:37

    \[\langle Jx,y\rangle =\langle x,Jy\rangle ,\quad \langle \langle J’x,y\rangle =\langle x,J’y\rangle .\] ->
    za dużo nawiasów kontowych

  6. Bjab says:
    2026/05/21 at 12:42

    caseof interest ->
    case of interest

  7. Bjab says:
    2026/05/21 at 12:43

    we have

    \[ \langle v,v\rangle =\langle JJ’x,J’Jv\rangle .\]

    x -> v

  8. Bjab says:
    2026/05/21 at 14:23

    Their norms in X_J are 1,1,\sqrt{2},\sqrt{2}

    Who cares about the norms of eigenvectors ? Scaled eigenvector is still eigenvector.

  9. Bjab says:
    2026/05/21 at 14:36

    “Eigenvectors of JJ’ corresponding to eigenvalues \lambda\neq 1 are always isotropic vectors in (X,\langle \cdot,\cdot\rangle ).”

    What if lambda = -1 ?

    • arkajad says:
      2026/05/21 at 16:22

      T is a positive operator on a Hilbert space. So all its eigenvalues are always strictly positive.

      It is instructive ro realize that eigenvectors are mutually orthogonal in both scalar products. These are two different scalar products. So their norms with respect to these two scalar products must differ. It is instructive to see how exactly they differ. But you are right: the orthogonal projections on eigenvector subspaces do not depend on the norms. They are the same operators in both Hilbert spaces. In general Hermitian operators for one scalar product will not be Hermitian for another scalar product. But for these particular projection operators we have an interesting exception – they are Hermitian in both scalar products.

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