Examples of Minkowski space

Question

Definition: The $n$-dimensional Minkowski Space, denoted by $M$, is the $n$-dimensional real vector space $\mathbb{R}^n$ with a bilinear form $g :\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ satisfying the following properties:

1. $\forall x,y \in \mathbb{R}^n$, $g(x,y)=g(y,x)$

2. If $g(x,y)=0, \forall y \in \mathbb{R}^n$, then $x=0$

3. $\exists$ basis $\{e_0,e_1,...,e_{n-1}\}$ for $\mathbb{R}^n$ with $g(e_i,e_j)= 1$ if $i=j=0$ and $g(e_i,e_j)= -1$ if $i=j=1,2,...,n-1$ and $g(e_i,e_j)= 0$ if $i\neq j$

This is the formal definition of Minkowski space.

Can anybody give examples of this? or how to construct such an example?

Answer 1

The properties tell you how to define the bilinear form $g$. Let me explain this for a completely general bilinear form. One can then easily see that the matrix $M$ one obtains is identical to the matrix in the answer of @Daron.

In general, given a basis $\{v_1,...,v_n\}$ for an $n$-dimensional vector space, and given a function $g$ which assigns to each pair of vectors $v_i,v_j$ a real number $a_{ij} = g(v_i,v_j)$, there is a unique way to extend $g$ to a bilinear form $g : V \times V \to \mathbb{R}$.

To define this extension, first form the matrix $M$ whose $ij^{\text{th}}$ entry is $a_{ij}$.

Notice that, so far, we have the special identity $$g(v_i,v_j) = v_i M v_j^T $$ where I assume by default that $v_i$ is represented by a row matrix with respect to the basis $\{v_1,....,v_n\}$, having $1$ in the $i^{\text{th}}$ entry and $0$'s elsewhere. Note that $v_j^T$ is a column matrix.

The special identity can then be extended to the general definition $$(*) \qquad\qquad g(v,w) = v M w^T \hphantom{\qquad\qquad(*)} $$ where, again, the vectors $v,w \in V$ are expanded as linear combinations of basis elements $$v = a_1 v_1 + \cdots + a_n v_n $$ $$w = b_1 v_1 + \cdots + b_n v_n $$ and are then represented as row matrices with the given coefficients; $$v = (a_1,...,a_n) $$ $$w = (b_1,...,b_n) $$ You can check, with a calculation, that if you evaluate the left hand side of $(*)$ by substituting the linear combination formulas for $v,w$ into $g(v,w)$ and applying the definition of a bilinear form, and if you evaluate the right hand side of $(*)$ by carrying out the matrix multiplication, the results are identical.

If one further requires the symmetry property (1) $g(v,w)=g(w,v)$, as for example in the Minkowsi form, then the real numbers $a_{ij}$ must themselves satisfy the symmetry property $a_{ij}=a_{ji}$, and one can then check again by a simple calculation that $g(v,w)=g(w,v)$ for all $v,w \in V$. The values of $a_{ij}$ given in (3), are evidently symmetric in this fashion.

Also, if one further requires the nondegeneracy property (2) then the matrix of real numbers $M = (a_{ij})$ must have nonzero determinant, and one can then check that (2) holds. The values of $a_{ij}$ of the Minkowsi form given in (3) evidently satisfy $\text{det}(M) = -1 \ne 0$.

Answer 2

Define the inner product by $(v,w) = v_0w_0 - \displaystyle \sum_{i>1}v_i w_i$.

All Minkowski spaces arise from an inner product of the form above, relative to some basis.

But if the vectors are not expressed under that basis the form of the inner product will not be apparent.

Expressed in an arbitrary basis, the inner product on a Minkowski space looks like $(v,w) = v A M A^t w^t $ where $A$ is some invertible matrix and. . .

$M= \left( \begin{array}{cccc} 1 & & & \\ & -1 & & \\ & & \ddots & \\ & & & -1\end{array} \right)$