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Let $(P, X, \pi, G)$ and $(P', X, \pi', G')$ be two principal bundles (with Lie groups $G$, $G'$ respectively). Given a vector space $V$ and representations $\rho, \rho'$ of the Lie groups in this vector space that they satisfy $\rho(g) \rho'(g') = \rho'(g') \rho(g)$ for $(g, g') \in G\times G'$ , consider the associated bundles $\xi = (P \times_{\rho} V, X, V)$ and $\xi' = (P' \times_{\rho'} V, X, V)$. If I make the "product bundle" (David Bleecker Gauge theory and variational principles page 90-91)

$$P \circ P' := \{(p, p') \in P \times P'| \pi(p) = \pi'(p')\},$$ will be the associated vector bundle

$$P \circ P' \times_{\rho \times \rho'} V$$ related to the tensor bundle $\xi \otimes \xi'$ in some way? If they are related, what is this relation?

I have this doubt because, in gauge theory, some authors use tensor product of vector bundles to get a final vector bundle and others using the "product bundle" mentioned above, and I want know if this things are equivalent, but I don't find this in none place.

Appreciate.

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    $\begingroup$ What is $\rho\times \rho'$? $\rho$ and $\rho'$ do not induce a representation of $G\times G'$ in $GL(V)$. $\endgroup$
    – Nicolast
    Commented May 26, 2021 at 11:58
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    $\begingroup$ Excuse me. I forgot to mention that they satisfy $\rho(g) \rho'(g') = \rho'(g') \rho(g)$ for $(g, g') \in G\times G'$. So just define $(g, g') \mapsto \rho(g) \rho'(g')$. This $\rho\times \rho'$ is more related with $\rho \otimes \rho'$, I put in this way because this is the notation of David Bleecker in his book "Gauge theory and variational principles" page 91. $\endgroup$ Commented May 26, 2021 at 12:29
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    $\begingroup$ That (still) doesn't make sense, the vector bundle you associate to $P\circ P'$ has rank $\dim V$, while the tensor product has rank $(\dim V)^2$. $\endgroup$
    – abx
    Commented May 26, 2021 at 12:55
  • $\begingroup$ You are right. I don't have consider this. I will change my question. Thank you abx. $\endgroup$ Commented May 26, 2021 at 14:11

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