Hi all,

Currently I'm reading a paper about the geometry of Grassmannians:


In there, the author regards the second fundamental form of the k-dimensional tautological bundle $\tau \rightarrow Gr:=Gr_k(\mathbb{C}^N)$ over a grassmannian given by

$H=\pi_{\tau^c} \circ \nabla^0 \in \Omega^1(Gr,Hom(\tau,\tau^c))$.

Here, $\tau^c$ denotes the complementary bundle of $\tau \subset Gr\times \mathbb{C}^N$, such that $\tau \oplus \tau^c \cong Gr \times \mathbb{C}^N$ with respect to the natural hermitian metric and $\pi_{\tau^c}$ is the projection on it. From this embedding we obtain natural connections on $\tau\rightarrow Gr$ and $\tau^c\rightarrow Gr$ by projecting the flat connection. It's easy to show that $H$ is indeed a 1-form with values in $Hom(\tau,\tau^c)$ as claimed.

In Lemma 2.1 on page 43 the author says that both the second fundamental form of $\tau\rightarrow Gr$ and $\tau^c\rightarrow Gr$ is parallel. Now at this point I have my question: As far as I know there is only one sense what parallel in this context means, namely

$(\nabla_X H)_Y\sigma:=\nabla^{\tau^c}_X(H_Y\sigma)-H_{\nabla_X Y}\sigma-H_Y\nabla^{\tau}_X\sigma=0$.

Can this be true? My first naive approach was writing the whole equation in terms of the flat connection and projections but this leads to nothing, I guess.

I wonder why none of the well known textbooks on differential geometry (of complex vector bundles) e.g. Kobayashi doesn't mention this fact if it is true. Maybe thd definition of "parallel" is not the one meant. So what could be the right one then? I would be grateful for any references.

best regards, Alex

  • 1
    $\begingroup$ While I haven't looked at the details, it sounds quite believable that H is parallel. In general, homogeneous things have constant curvatures... $\endgroup$ Commented Aug 23, 2012 at 22:08
  • $\begingroup$ Yes, using theory about the holonomy group of homogeneous spaces together with the existence of invariant connections on homogeneous spaces seems to lead to an answer. I've not yet figured out the details but: I still wonder about the fact why there isn't a simple direct calculation. $\endgroup$
    – Alex_K
    Commented Aug 24, 2012 at 16:17

1 Answer 1


Don't know if anyone is interested in the (quite simple) answer: The bundles $Hom(\tau,\tau^c)$ and $TGr$ are not only isomorphic as bundles but also as bundles with their natural connection, i.e. the induced Hom-connection from $\nabla^{\tau}$ respectively $\tau^{\tau^c}$ and den Levi-Civita-Connection. We have to show: $\nabla_X(H_Y\sigma)=H_{\nabla_XY}\sigma+H_Y\nabla_X\sigma$ where everytime the only possible connection is meant. This is equivalent to $\nabla_XH_Y\sigma-H_Y\nabla_X\sigma=H_{\nabla_XY}\sigma$ but $\nabla_XH_Y\sigma-H_Y\nabla_X\sigma=:(\nabla^{Hom}_XH_Y)(\sigma)$. Using the fact, that the isomorphism between $Hom(\tau,\tau^c)$ and $TGr$ is given by $X \mapsto H_X$ and the fact, that this isomorphism preserves the connection, we're done.

You could also use the fact, that $Gr=G/H$ is a homogeneous space, so $G(G/H,H)$ is a principle $H$-fibre bundle with left $G$-action and there is a $G$-invariant connection on this fibre bundle. The $G$-invariance of all tensor fields gives then also that the second fundamental form is parallel.


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