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Date: Mon, 10 May 2004 21:29:01 +0200 (CEST)
Subject: Second Tex File
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The vector bundle $E \otimes E^\ast$ is obviously defined pointwise, that
is $(E \otimes E^\ast)_x = E_x \otimes (E_x)^\ast$. We know that for
finite vector spaces, $E_x \otimes (E_x)^\ast$ is isomorphic to Hom$(E,
E)$, and since $E$ is 1-dimensional, Hom$(E,E) \cong R$. Denote this
isomorphism by $\varphi_{x}$.

\bigskip\noindent
Then we define the natural isomorphism $\varphi : E \otimes E^\ast
\longrightarrow X \times R$ by $\varphi(e \otimes v) := (p(e),
\varphi_{p(e)}(e \otimes v)$
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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 $g^{E \otimes F}_{ij} = g^E_{ij} \cdot g^F_{ij}$

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The vector bundle $E \otimes E^\ast$ is obviously defined pointwise, that is $(E \otimes E^\ast)_x = E_x \otimes (E_x)^\ast$. We know that for finite vector spaces, $E_x \otimes (E_x)^\ast$ is isomorphic to Hom$(E, E)$, and since $E$ is 1-dimensional, Hom$(E,E) \cong R$. Denote this isomorphism by $\varphi_{x}$.

\bigskip\noindent
Then we define the natural isomorphism $\varphi : E \otimes E^\ast \longrightarrow X \times R$ by $\varphi(e \otimes v) := (p(e), \varphi_{p(e)}(e \otimes v)$
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