Subject: My Tex Thing 1
From: [email-masked]
To: [email-masked]
Date: Mon, 10 May 2004 21:07:02 +0200 (CEST)

If $E_x$ and $F_x$ are 1-dimensional, the isomorphisms $\phi_{i,x}: E_x
\longrightarrow R$ and $\psi_{i,x}: F_x \longrightarrow R$ can be
represented by a $1 \times 1$ matrix, i.e. \emph{multiplication by a real
number}. Let us denote this number with $A_{i,x}$ and $B_{i,x}$ for
$\phi_{i,x}$ resp. $\psi_{i,x}$. Then, obviously $g^E_{ij}(x) =
\frac{A_{i,x}}{A_{j,x}$ and $g^F_{ij}(x) = \frac{B_{i,x}}{B_{j,x}$,
$\forall x \in X$.

\bigskip\noindent
The tensor product of vector bundles is defined pointwise, i.e. $(E
\otimes F)_x = E_x \otimes F_x \; \forall x \in X$. Therefore the
isomorphisms which we get are $\chi_{i,x} : E_x \otimes F_x
\longrightarrow R \otimes R \; \cong \; R$ given by $\chi(e \otimes f) :=
\phi_{i,x}(e) \cdot \psi{i,x}(f) = A_{i,x} \cdot B_{i,x} \cdot e \cdot f$.
It's inverse is then given by $\chi^{-1}_{i,x}(c) := \frac{c}{A_{i,x}
\cdot B_{i,x}}.

\bigskip\noindent
It follows that for every $x \in R$ the transition maps are given by
$$
g^{E \otimes F}_{ij}(x) := (\chi_{i,x} \circ \chi_{j,x}^{-1})(x) = [c
\longmapsto \frac{c \cdot A_{i,x} \cdot B_{i,x}}{A_{j,x} \cdot B_{j,x}} =
c \cdot g^E_{ij}(x) \cdot g^F_{ij}(x)]
$$
Hence $g^{E \otimes F}_{ij}(x)$ is multiplication by the real number
$g^E_{ij}(x) \cdot g^F_{ij}(x)$ $\forall x \in X$, and hence $g^{E \otimes
F}_{ij} = g^E_{ij} \cdot g^F_{ij}$