# Notation regarding generalized Minkowski space

In section 12 of the book Surfaces in classical geometries: A treatment by moving frames by Gary R. Jensen, Emilio Musso and Lorenzo Nicolodi (see preview here), Möbius geometry is described.

They introduce the generalized Minkowski space of signature $$(4,1)$$ as follows:

Let $$R^{4,1}$$ denote $$R^5$$ with a Lorentzian inner product. Let $$\epsilon_0,\dots,\epsilon_4$$ denote the standard orthonormal basis of $$R^{4,1}$$ given by the standard orthonormal basis $$\epsilon_0,\dots,\epsilon_3$$ of the Euclidean space $$R^4$$ and with $$\langle \epsilon_4, \epsilon_4\rangle=-1$$. The Lorentzian inner products $$\langle \epsilon_a, \epsilon_b\rangle$$, for $$a,b=0,\dots,4$$, are the entries of the matrix $$\begin{pmatrix} I_4 & 0 \\ 0 & -1 \end{pmatrix}.$$ Write elements of $$R^{4,1}=R^4\oplus R\epsilon_4$$ as $$x + t\epsilon_4$$, where $$x\in R^4$$ and $$t\in R$$. The Lorentzian inner product is then $$\langle x+s\epsilon_4,y+t\epsilon_4\rangle=x\cdot y-st.$$

My problem is that I would expect vectors in $$R^{4,1}$$ to have dimension $$5$$, just like vectors in the Minkowski space have dimension $$4$$. They actually start stating that $$R^{4,1}$$ is $$R^5$$ with a Lorentzian inner product. This could suggest that it is just a typo (two actually), but later on in the chapter they are consistent on this matter: vectors in $$R^{4,1}$$ have $$4$$ components. For instance, they consider a mapping which is the sum of a vector in $$\mathbb{S}^3$$ with the vector $$\epsilon_4$$. Therefore I assume there's no typo and it's just my poor understanding...

Can anyone help understanding this construction and how it works?

• I'm confused. Where exactly do you see a typo or the statement that elements of $R^{4,1}$ have $4$ components? – freakish Oct 13 '18 at 13:33
• Note the four vectors $\epsilon_0,\epsilon_1,\epsilon_2,\epsilon_3$ give the +ve subspace and $\epsilon_4$ is -ve subspace of the inner product. The slightly nonstandard way of indexing the +ve subspace probably caused your confusion. (By the way, the usual definition of signature is not going to give (5,1) but (4,1)). – user10354138 Oct 13 '18 at 13:45
• @freakish They write vectors of $R^{4,1}$ as $x+t\epsilon_4$, where $x\in R^4$. This plus the fact that the $\epsilon_i$ for $i=0,\dots,3$ are the standard orthonormal basis of $R^4$ makes me think that $x+t\epsilon_4$ is a sum of $4$-vectors... What am I getting wrong? – Edu Oct 13 '18 at 14:14
• @user10354138 That's clear but again, the say that $\epsilon_0,\dots,\epsilon_3$ is the standard orthonormal basis of $R^{4}$... how is that they don't have $4$ components then? – Edu Oct 13 '18 at 14:16
• @Edu This is not a sum of $4$-vectors. First of all note that $e_4\in R^5$. What they say is that because $R^{4,1}=R^4\oplus R$ then every 5-vector can be written as $x+te_4$ where $x\in R^4\oplus 0$ and $e_4\in 0^4\oplus R$ in a unique way. When they write $x\in R^4$ what they actually mean is that $x$ is in a $4$-dimensional subspace of $R^5$, i.e. formally $x\in R^4\oplus 0$ meaning $x=(x_1,x_2,x_3,x_4, 0)$. In other words they say that any 5-vector can be written as $(x_1,x_2,x_3,x_4,0)+t(0,0,0,0,1)$ which is quite obvious. – freakish Oct 13 '18 at 14:27

So when they write

Write elements of $$R^{4,1}=R^4\oplus Re_4$$ as $$x+te_4$$, where $$x\in R^4$$ and $$t\in R$$.

what they actually mean is that $$x$$ is in the $$4$$-dimensional subspace of $$R^{4,1}$$ spanned by $$\{e_1,e_2,e_3,e_4\}$$. It doesn't mean that $$x\in R^4$$ literally, $$x$$ is still an element of a $$5$$-dimensional space but one of the coordinates is zerod.

For example if we take the standard basis $$e_i=(0,\ldots,0,1,0,\ldots, 0)$$ with $$1$$ at $$i$$-th posisition then the statement can be rewritten as

Write elements of $$R^{4,1}$$ as $$(x_0,x_1,x_2,x_3,0)+t(0,0,0,0,1)$$.

More generally the precise statement should be

Write elements of $$R^{4,1}$$ as $$x+te_4$$ where $$x=x_0e_0+x_1e_1+x_2e_2+x_3e_3$$ for $$x_0,x_1,x_2,x_3,t\in R$$.

Here $$R^4$$ means $$span(e_0,e_1,e_2,e_3)$$. It is a $$4$$-dimensional subspace of $$R^5$$.