5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Isn't this closer to algebraic topology than differential geometry? $\endgroup$ – Will Sawin Jun 10 '17 at 12:57
  • 2
    $\begingroup$ 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 Jun 10 '17 at 13:33
  • 1
    $\begingroup$ @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 Jun 10 '17 at 14:02
  • 1
    $\begingroup$ 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 Jun 10 '17 at 23:35
5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jun 10 '17 at 15:12
  • 1
    $\begingroup$ @JasonStarr What representation of $PGL_2(\mathbb R)$ will you take? Surely not the adjoint representation... $\endgroup$ – Will Sawin Jun 10 '17 at 15:17
  • $\begingroup$ @JasonStarr Also I forgot to mention $G$ should be compact. $\endgroup$ – Will Sawin Jun 10 '17 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 Jun 10 '17 at 15:20
  • 1
    $\begingroup$ @JasonStarr But the rank 4 faithful representation is adjoint + trivial, so if that is nontrivial then the adjoint is trivial. $\endgroup$ – Will Sawin Jun 10 '17 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.