I am looking for an example of the following situation. Let $M$ be a connected (if possible compact) manifold such that its tangent bundle $T(M)$ admits a vector bundle decomposition $$ T(M) = A \oplus L_1 \oplus \dotsb \oplus L_k, ~~~ k > 0, $$ where each summand $L_i$ is a line bundle over $M$, and $A$ is a vector bundle that cannot be written as a direct sum of line bundles. If $M$ admitted a nice (compact) group action then that would be even better.
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2$\begingroup$ Is K supposed to be L? Also, should I assume k>0? $\endgroup$– Sean SanfordCommented Jul 5 at 18:29
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$\begingroup$ yes, it is now fixed. $\endgroup$– Bobby-John WilsonCommented Jul 5 at 19:25
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$\begingroup$ I guess in your question you implicitly want that the splitting is preserved by group action. yes? $\endgroup$– Ali TaghaviCommented Jul 22 at 9:46
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3 Answers
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You are asking about manifolds whose projective span is equal to $k$, according to the main definition of Grant and Schutte - Projective span of Wall manifolds.
Any manifold with zero Euler characteristic gives an example with $k>0$, but you'll find many more interesting examples in the cited preprint.
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Remark: I assume that you want $A$ to be a non-trivial bundle. Otherwise, of course, any parallelizable compact manifold would be an example. In particular, any compact Lie group would be an example with a nice compact group action.
A simple interesting example is $S^5$, whose tangent bundle can be written as the sum of a line bundle and an irreducible 4-plane bundle. It also has a nice compact group action, as $\mathrm{SO}(6)$ acts transitively on it in the obvious way.
The proof of this is as follows: First, $S^5$ is an odd dimensional sphere, so it has a nonvanishing tangent vector field. Hence $TS^5 = L_1 \oplus A$ where $L_1$ is a trivial line bundle and $A$ is a 4-plane bundle. It remains to show that $A$ is irreducible. First, $A$ cannot contain any non-trivial line bundle $L_2$ since $L_2$ would have to be orientable (since $S^5$ is simply connected) and hence trivial, and a non-vanishing section of such an $L_2$ would yield a tangent vector field on $S^5$ that is linearly independent from any non-vanishing section of $L_1$. However, as is well-known, $S^5$ cannot have two linearly independent tangent vector fields. Second, $A$ cannot split as the sum of two $2$-plane bundles since $H^2(S^5,Z)$ is trivial, so any $2$-plane bundle over $S^5$ must be trivial and hence a sum of two (trivial) line bundles. Hence, if $A$ could be written as the sum of two $2$-plane bundles, $S^5$ would have to be parallelizable, which, we know, it is not. Thus, $A$ is irreducible.
answered Jul 7 at 16:46
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$\begingroup$ However the tangent bundle of $S^5$ is not decompossible in the way you mentioned but according to the Splitting principal there is an space $X$ and continuous map $f:X\to S^5$ such that $f^*$ is an isomorphism in K theory and the pull back $f^*(TS^5)$ of the tangent bundle of $S^5$ is decomposed to direct sum of line bundle. In this case s there an explicit (compact) manifold $X$ with this property?What is theminimum possible dimension for such manifold $X$? By this question I search for a possible smooth version of splitting principle in K theory $\endgroup$ Commented Jul 21 at 7:17
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$\begingroup$ I just ask this question and its extension here mathoverflow.net/questions/475457/… $\endgroup$ Commented Jul 21 at 9:16
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$\begingroup$ In this example I think the plane $A$ is the pull back of the real tangent bundle of $\mathbb{C}P^2$ under the fiber map $S^5 \to \mathbb{C}P^2$. The tangent bundle of CP^2 is irreducible but it apparently does not implies that the bundle A you mentioned is irreducible. So this is a motivation to ask: What is an example of principle bundle $X\to B$ where $A$ is a connection plane tangent to X such that $TB$ is irreducible but A is not? I would appreciate if you give comment on this question $\endgroup$ Commented Jul 22 at 15:55
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1$\begingroup$ @AliTaghavi: A simple example, is to let $X = S^3$ and $B=S^2$ and let $\pi:S^3\to S^2$ be the Hopf map. The tangent bundle to $S^2$ is irreducible over $S^2$, but the pullback via $\pi$ is trivial over $S^3$. $\endgroup$ Commented Jul 22 at 17:43
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$\begingroup$ Thank you very much for your answer $\endgroup$ Commented Jul 22 at 21:16
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The product of a 2-sphere with a torus is an example. There are complex analytic examples due to Beauville (Complex manifolds with split tangent bundle). Every 3-manifold has trivial tangent bundle, so we could use a product of a torus, a 3-manifold, and a 2-sphere.
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1$\begingroup$ Doesnt the sphere times torus have a trivial tangent bundle? $\endgroup$ Commented Jul 6 at 11:26
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1$\begingroup$ @ThomasRot: yes, sorry, that's right. But Beauville's examples are nice complex analytic examples. $\endgroup$ Commented Jul 6 at 13:51