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Let $P$ be be a principal bundle over a manifold $M$ with structure group $G$, where $G$ is a Lie group. Let $E = P\times_{\rho} \mathbb{R}^{k}$ be a vector bundle associated to $P$ through a faithful representation $\rho\colon G\to Gl(k,\mathbb{R})$. Let $\mathrm{Ad}(P) = P\times_{Ad}\mathfrak{g}$ denote the adjoint bundle associated to $P$, where $\mathfrak{g}$ denotes the Lie algebra of $G$. My question is the following:

If $\mathrm{Ad}(P)$ is topologically trivial, namely $\mathrm{Ad}(P)\simeq M\times \mathfrak{g}$, does it follow then that the endomorphism bundle $\mathrm{End}(E)\simeq E\otimes E^{\ast}$ of $E$ is also topologically trivial?

If I am not mistaken, this is clearly true if $G=Gl(k,\mathbb{R})$. I am interested in the case $G\subset Gl(k,\mathbb{R})$ and $G\neq Gl(k,\mathbb{R})$, with $G$ connected, compact and semi-simple.

Thanks.

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    $\begingroup$ Isn't this closer to algebraic topology than differential geometry? $\endgroup$
    – Will Sawin
    Commented Jun 10, 2017 at 12:57
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    $\begingroup$ What if $G$ is a finite group, say $\mathbb{Z}/n\mathbb{Z}$? Then the Lie algebra $\mathfrak{g}$ is the zero-dimensional vector space, so that $\text{Ad}(P)$ is necessarily trivial. Yet the vector bundle $E$ may be nontrivial, e.g., the orientation bundle of a non-oriented manifold. $\endgroup$
    – Jason Starr
    Commented Jun 10, 2017 at 13:33
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    $\begingroup$ @JasonStarr Note that I am wondering about the relation between the triviality of $\mathrm{Ad}(P)$ and the triviality of $\mathrm{End}(E)$, not $E$. Isn't the endomorphism bundle of any real line bundle trivial? (trivialized by the identity section). $\endgroup$
    – Bilateral
    Commented Jun 10, 2017 at 14:02
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    $\begingroup$ Here is an idea. Choose $M$ to be an $m$-sphere with $m\cong 3,5,6,7 \ (\text{mod}\ 8)$ such that the rank $n$ of $\mathfrak{g}$ satisfies $n>m$. Then by Bott periodicity, $\text{Ad}(P)$ is trivial. Yet the rank of $E$ might be small enough so that $E$, and even $\text{End}(E)$, are nontrivial. $\endgroup$
    – Jason Starr
    Commented Jun 10, 2017 at 23:35

2 Answers 2

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This was too long for a comment. There are counterexamples where $G$ is a finite group. The OP correctly notes that the endomorphism bundle of a rank $1$ bundle is trivial. Nonetheless, there are plenty of examples in higher rank with nontrivial endomorphism bundle.

For instance, let $G$ be the cyclic group of order $2$. Begin with the real projective plane $\mathbf{RP}^2 = \mathbf{S}^2/G$ with its tautological $G$-bundle, $\mathbf{S}\to \mathbf{RP}^2$. The standard inclusion $G\subset \textbf{GL}_1(\mathbb{R}) = \mathbb{R}^\times$ induces the tautological rank $1$ bundle $\gamma_2$. Let $M$ be a product $X\times Y$ of two real projective planes $X\cong Y\cong \mathbf{RP}^2 = \mathbf{S}^2/G$. The $G\times G$-bundle $$\widetilde{X}\times \widetilde{Y}\to X\times Y$$ and the inclusion $$G\times G\subset \mathbf{GL}_1(\mathbb{R})\times \mathbf{GL}_1(\mathbb{R}) \subset \mathbf{GL}_2(\mathbb{R})$$ induces a rank $2$ bundle $E$ that is simply $\text{pr}_X^*\gamma_2 \oplus \text{pr}_Y^*\gamma_2$.

The endomorphism bundle is isomorphic to a rank $4$ bundle, $$\text{End}(E) \cong \mathbb{R} \oplus (\text{pr}_X^*\gamma_2\otimes \text{pr}_Y^*\gamma_2^\vee) \oplus (\text{pr}_X^*\gamma_2^\vee \otimes \text{pr}_Y^*\gamma_2) \oplus \mathbb{R}.$$ By the Whitney sum formula, the total Stiefel-Whitney class is $$(1)(1+\text{pr}_X^*a - \text{pr}_Y^*a)(1-\text{pr}_X^*a+\text{pr}_Y^*a)(1) =$$ $$ 1 - (\text{pr}_X^*a)^2 - (\text{pr}_Y^*a)^2 + 2\text{pr}_X^*a\cup \text{pr}_Y^*a.$$ Of course the last summand is zero since the coefficients are $\mathbb{Z}/2\mathbb{Z}$. However, $a^2$ is nonzero in $H^2(\mathbb{RP}^2;\mathbb{Z}/2\mathbb{Z})$. Using Künneth, the second Stiefel-Whitney class of $\text{End}(E)$ is nonzero.

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There are counterexamples for $G$ connected abelian.

Let $X = \mathbb CP_2$, $P$ be the complex line bundle $\mathcal O(1)$, viewed as a $G=SO(2)$-bundle. We take $\rho$ the standard two-dimensional representation of $SO(2)$. Of course the adjoint bundle is trivial in this case.

To show that $E \otimes E^*$ is nontrivial, we calculate its first Pontryagin class by tensoring with $\mathbb C$, where we can view it as $(\mathcal O(1) + \mathcal O(-1)) \otimes ( \mathcal O(-1) + \mathcal O(2) ) = \mathcal O(2) + \mathcal O + \mathcal O + \mathcal O(-2)$, which has second chern class $-4 H^2$, which is nontrivial, hence is a nontrivial complex vector bundle (and thus $E \otimes E^*$ is a nontrivial real vector bundle).

However, the statement might be possibly be true for $G$ connected, semisimple, and compact.

Let $Z$ be the center of $G$. Then the condition that $Ad(P)$ is trivial necessarily implies that the $G/Z$ bundle on $M$ induced by $P$ is pulled back from the obvious $G/Z$-bundle on $Gl_n(\mathfrak g)/ (G/Z)$, where $G/Z$ acts by right multiplication in the adjoint representation.

If $\rho$ is a representation on which $Z$ acts by scalars, then $E \otimes E^*$, as a representation of $G$, factors through $G/Z$, and hence is a pullback from $M$.

So in this special case, it suffices to check whether the vector bundle $E \otimes E^*$ on $GL_n(\mathfrak g) / (G/Z)$ is nontrivial. However, I don't know how to do this.

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  • $\begingroup$ I suspect there are counterexamples with $G$ connected semisimple that are induced from a disconnected group $\Gamma$. For instance, for $G=\textbf{PGL}_2(\mathbb{R})$ and $\Gamma$ the subgroup $\cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ generated by the following matrices, $$\left[ \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right], \ \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right].$$ $\endgroup$
    – Jason Starr
    Commented Jun 10, 2017 at 15:12
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    $\begingroup$ @JasonStarr What representation of $PGL_2(\mathbb R)$ will you take? Surely not the adjoint representation... $\endgroup$
    – Will Sawin
    Commented Jun 10, 2017 at 15:17
  • $\begingroup$ @JasonStarr Also I forgot to mention $G$ should be compact. $\endgroup$
    – Will Sawin
    Commented Jun 10, 2017 at 15:19
  • $\begingroup$ Perform the construction from my answer for $\mathbf{RP}^8 \cong \mathbf{S}^8/(\mathbb{Z}/2\mathbb{Z})$ rather than $\mathbf{RP}^2$. Then the "standard" rank $4$ faithful representation gives a vector bundle $F$ on $\mathbf{RP}^8\times \mathbf{RP}^8$ that is isomorphic to $\text{End}(E)$ from my answer. Thus, the degree $8$ Stiefel-Whitney class of $\text{End}(F)$ is $(\text{pr}_X^*a - \text{pr}_Y^*a)^8 = \text{pr}_X^*(a^8) + \text{pr}_Y^*(a^8)$ by the Whitney sum formula. $\endgroup$
    – Jason Starr
    Commented Jun 10, 2017 at 15:20
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    $\begingroup$ @JasonStarr But the rank 4 faithful representation is adjoint + trivial, so if that is nontrivial then the adjoint is trivial. $\endgroup$
    – Will Sawin
    Commented Jun 10, 2017 at 15:21

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