- #1
mma
- 270
- 5
- TL;DR
- What is this beyond a calculation trick to calculate the pseudo-norm as if it were an Euclidean norm?
In relativity theory, it's a common habit to use a quadruplet x=(x_0, x_1, x_2, x_3) with x_0=ict (or with c=1, x_0=it ) instead of (t,x_1,x_2,x_3)\in \mathbb R^4, and to use the formal Euclidean metric \|x\|^2=\sum_{i=0}^3x_i^2 instead of the Minkowski pseudo-metric -t^2 + \sum_{i=1}^3x_i^2. But what is (it,x_1,x_2,x_3) by itself? For a long time I didn't think there was any real mathematics behind this formal computation trick. Until I came across F. F. Eberlein's "The Spin Model of Euclidean 3-space"1 and "Models of spacetime"2. From them I learned that it is possible to make mathematically well-defined sense of this.
Eberlein's Euclidean 3-space \mathfrak E_3 consists of the traceless, Hermitian 2\times 2 complex matrices with scalar product A\cdot B:=\frac{1}{2}(AB+BA)/I where AB means the product of the matrices A and B, I is the 2\times 2 identity matrix and for a diagonal matrix D=\lambda I, D/I:=\lambda. Eberlein proves that for any A, B\in\mathfrak E_3, AB is always diagonal, (AB+BA)/I is real-valued, and the above-defined \mathfrak E_3^2\to \mathbb R: (A,B)\mapsto A\cdot B function is a symmetric, positive definite nondegenerate bilinear form, i.e. and Euclidean scalar product. In \mathfrak E_3, the Pauli-matrices form an orthogonal basis. Furthermore, A\cdot A=-\mathrm{det}A, and i\mathfrak E_3=\mathfrak{su}(2)
The space of quaternions, \mathbb H:=\mathbb RI\oplus\mathfrak{su}(2) = \mathbb RI + i\mathfrak E_3 is also an Euclidean space with Euclidean norm \|q\|^2=\mathrm{det}(q). For q = tI + iA\in\mathbb H (A\in\mathfrak E_3), \mathrm{det}(q) = t^2 - \mathrm{det}(A), so, taking an orthonormal basis (\xi_1, \xi_2, \xi_3) in \mathfrak E_3 and taking any q = tI + i\sum_{i=1}^nx_i \xi_i\in \mathbb H (t,x_i\in\mathbb R), \|q\|^2=t^2+\sum_{i=1}^3x_i^2.
In contrast of this, the space \mathbb RI \oplus \mathfrak E_3 with the pseudo-norm \|v\|^2=\mathrm{det}(v) is a Minkowski space with signature (1,3), that is, the space M:=i\mathbb RI \oplus i\mathfrak E_3=i\mathbb RI \oplus \mathfrak{su}(2) with pseudo-norm \|v\|^2 = \mathrm{det}(v) is a Minkowski space with the usual signature (3,1). Taking any v = itI + i\sum_{i=1}^nx_i \xi_i\in M (t,x_i\in\mathbb R, (\xi_1, \xi_2, \xi_3) is an orthonormal basis in \mathfrak E_3), \|v\|^2=(it)^2+\sum_{i=1}^3x_i^2
Now, if we take an orthonormal basis \{I,e_1,e_2,e_3\} in the Euclidean space \mathbb H, and take a vector (t,x_1,x_2,x_3):=tI+\sum_{i=1}^3x_ie_i, then (it,x_1,x_2,x_3):=itI+\sum_{i=1}^3x_ie_i is a vector in the Minkowski space M. In short, (it,x_1,x_2,x_3) are the coordinates of a vector of M with respect to an orthonormal basis of \mathbb H. This makes sense, since both \mathbb H and \mathbb M consist of 2\times 2 complex matrices.
The above mathematization of (it,x_2,x_2,x_3) heavily relies on the specific model of the Euclidean space and the Minkowski space. I wonder if is there another way to make (it,x_2,x_2,x_3) mathematically meaningful?
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1Am. Math. Mon. 69, 587-598 (1962). A crucial correction. Ibid. 69, 960 (1963).
2Bull. Amer. Math. Soc. 71 (1965), 731-736
Eberlein's Euclidean 3-space \mathfrak E_3 consists of the traceless, Hermitian 2\times 2 complex matrices with scalar product A\cdot B:=\frac{1}{2}(AB+BA)/I where AB means the product of the matrices A and B, I is the 2\times 2 identity matrix and for a diagonal matrix D=\lambda I, D/I:=\lambda. Eberlein proves that for any A, B\in\mathfrak E_3, AB is always diagonal, (AB+BA)/I is real-valued, and the above-defined \mathfrak E_3^2\to \mathbb R: (A,B)\mapsto A\cdot B function is a symmetric, positive definite nondegenerate bilinear form, i.e. and Euclidean scalar product. In \mathfrak E_3, the Pauli-matrices form an orthogonal basis. Furthermore, A\cdot A=-\mathrm{det}A, and i\mathfrak E_3=\mathfrak{su}(2)
The space of quaternions, \mathbb H:=\mathbb RI\oplus\mathfrak{su}(2) = \mathbb RI + i\mathfrak E_3 is also an Euclidean space with Euclidean norm \|q\|^2=\mathrm{det}(q). For q = tI + iA\in\mathbb H (A\in\mathfrak E_3), \mathrm{det}(q) = t^2 - \mathrm{det}(A), so, taking an orthonormal basis (\xi_1, \xi_2, \xi_3) in \mathfrak E_3 and taking any q = tI + i\sum_{i=1}^nx_i \xi_i\in \mathbb H (t,x_i\in\mathbb R), \|q\|^2=t^2+\sum_{i=1}^3x_i^2.
In contrast of this, the space \mathbb RI \oplus \mathfrak E_3 with the pseudo-norm \|v\|^2=\mathrm{det}(v) is a Minkowski space with signature (1,3), that is, the space M:=i\mathbb RI \oplus i\mathfrak E_3=i\mathbb RI \oplus \mathfrak{su}(2) with pseudo-norm \|v\|^2 = \mathrm{det}(v) is a Minkowski space with the usual signature (3,1). Taking any v = itI + i\sum_{i=1}^nx_i \xi_i\in M (t,x_i\in\mathbb R, (\xi_1, \xi_2, \xi_3) is an orthonormal basis in \mathfrak E_3), \|v\|^2=(it)^2+\sum_{i=1}^3x_i^2
Now, if we take an orthonormal basis \{I,e_1,e_2,e_3\} in the Euclidean space \mathbb H, and take a vector (t,x_1,x_2,x_3):=tI+\sum_{i=1}^3x_ie_i, then (it,x_1,x_2,x_3):=itI+\sum_{i=1}^3x_ie_i is a vector in the Minkowski space M. In short, (it,x_1,x_2,x_3) are the coordinates of a vector of M with respect to an orthonormal basis of \mathbb H. This makes sense, since both \mathbb H and \mathbb M consist of 2\times 2 complex matrices.
The above mathematization of (it,x_2,x_2,x_3) heavily relies on the specific model of the Euclidean space and the Minkowski space. I wonder if is there another way to make (it,x_2,x_2,x_3) mathematically meaningful?
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1Am. Math. Mon. 69, 587-598 (1962). A crucial correction. Ibid. 69, 960 (1963).
2Bull. Amer. Math. Soc. 71 (1965), 731-736