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?

  • $\begingroup$ I'm confused. Where exactly do you see a typo or the statement that elements of $R^{4,1}$ have $4$ components? $\endgroup$ – freakish Oct 13 '18 at 13:33
  • $\begingroup$ 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)). $\endgroup$ – user10354138 Oct 13 '18 at 13:45
  • $\begingroup$ @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? $\endgroup$ – Edu Oct 13 '18 at 14:14
  • $\begingroup$ @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? $\endgroup$ – Edu Oct 13 '18 at 14:16
  • 1
    $\begingroup$ @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. $\endgroup$ – 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 $R^5$.


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