1
$\begingroup$

Inspried by this MO question A manifold whose tangent space is a sum of line bundles and higher rank vector bundles we pose the following question as a possible smooth version of the splitting principle in k theory.

Recall that the splitting principle says the following(Page 80 of "Vector Bundles and K theory")

Splitting Principle: "Let $E\to X$ be a vector bundle on a compact Hausdorff space $X$, then there is a compact space $Y$ and a continuous function $f:Y\to X$ such that $f^*$ is injective in K theory and $\mathbb{Z}/2$ cohomology. Moreover $f^*(E)$ is a direct sum of line bundles.

Is this principle valid in the smooth context? Can $Y$ be choosen a compact manifold of the same dimension as $X$ and $f$ a smooth map? What about the real analytic case?

Regardless of injectivity of $f^*$:

Is there a compact manifold $X$, preferably orientable which does not admit any smooth surjection $f:Y\to X$ where $Y$ is a comact parralelizable manifold?

Improve this question
$\endgroup$
21
  • 5
    $\begingroup$ The splitting principle constructs the space as a flag bundle. If $X$ is a compact smooth manifold, $Y$ will be as well, and $f$ will be a smooth map. I don't know anything about analyticity as I never thing of such things, but my gut says that it is also true in that context. The $Y$ will be of higher dimension. $\endgroup$
    – Thomas Rot
    Commented Jul 21 at 11:09
  • $\begingroup$ @abx Please read my last question again. The injectivity of $f^*$ is no longer assumed. moreover do you mean $T_X$ instead of $T_Y$? Note that $Y$ is parallelizable. So I do not get your comment. $\endgroup$
    – Ali Taghavi
    Commented Jul 21 at 13:16
  • $\begingroup$ @ThomasRot Analytic objects are discussed in Hirsch Differential topology(For upgrading smooth structures to analytic one) and also Kobayashi and Nomizu Differential geometry. Some times passing from smooth to analytic would creat interesting advantures. I know splitting principal in the CW complex setting so please ellaborate your comment in smooth case. I would appreciate if you provide an answer. $\endgroup$
    – Ali Taghavi
    Commented Jul 21 at 13:21
  • 1
    $\begingroup$ For the last question: The torus surjects to any connected compact manifold of the same (or lower) dimension $\endgroup$
    – Thomas Rot
    Commented Jul 21 at 21:48
  • 1
    $\begingroup$ @AliTaghavi Sorry, I do not understand the exercise that you sited. I think that $S^1$ is a union of two copies of open intervals, but it is not even homotopy equivalent to $\mathbb R^1$. $\endgroup$
    – Z. M
    Commented Jul 22 at 17:11

0

Reset to default

You must log in to answer this question.