Fix a vector space $V$ and an integer $1\leq n<\dim V$.

If $\mathcal{I}\subseteq\Lambda^\bullet V^*$ is an ideal, I denote by $\mathcal{I}^i:=\mathcal{I}\cap\Lambda^iV^*$ its $i^\textrm{th}$ homogeneous piece, and by $$ \ker\mathcal{I}^i:=\bigcap_{\omega\in \mathcal{I}^i}\ker\omega\subset\Lambda^iV $$ the linear subspace of $\Lambda^iV$ made by the elements which vanish when contracted with all the $i$-forms from $\mathcal{I}$.

Denote by $$ K(\mathcal{I}):=\prod_{i=1}^n\mathbb{P}(\ker \mathcal{I}^i)\subset\prod_{i=1}^n\mathbb{P}(\Lambda^iV)\subset\mathbb{P}\left( \bigotimes_{i=1}^n\Lambda^iV\right) $$ the Segre-image of the product of the projectivised kernels above (the reason why I start from $i=1$ is that I'm assuming $\mathcal{I}^0=0$).

Finally, I need the flag variety $$ \mathrm{Fl}(1,\ldots,n,V)\subset\mathbb{P}\left( \bigotimes_{i=1}^n\Lambda^iV\right) $$ the embedding being given by $$ (\langle v_1\rangle,\langle v_1, v_2\rangle,\ldots,\langle v_1,v_2,\ldots,v_n\rangle)\mapsto [v_1\otimes(v_1\wedge v_2)\otimes\cdots\otimes(v_1\wedge v_2\wedge\cdots\wedge v_n)]\, . $$ These two ingredients allow me to define the flags of integral elements $$ \mathrm{Fl}_n(\mathcal{I}):=K(\mathcal{I})\cap\mathrm{Fl}(1,\ldots,n,V)\, . $$

Motivation. In the theory of Exterior Differential Systems, if $V=T_xM$, then an element $(V_1,V_2,\ldots, V_n)$ of $\mathrm{Fl}_n(\mathcal{I})$ is called an integral flag at $x\in M$. If such a flag is a ``good one'', then one can iteratively use the Cauchy-Kowalewskaya theorem to construct the germ at $x$ of an integral submanifold of $V_n$.

Define now a sub-variety $$ \mathrm{Fl}^{\textrm{ord}}_n(\mathcal{I})\subseteq\mathrm{Fl}_n(\mathcal{I})\, , $$ by requiring it to be the largest one on which all the natural bundle $$ \mathrm{Fl}(1,\ldots,n,V)\longrightarrow \mathrm{Gr}(i,n)\, ,\quad i=1,\ldots,n\, ,\quad (^*) $$ restrict to sub-bundles (by sub-bundle I mean that both the fibre and the base can get smaller).

Motivation. The "good flags" above are called "ordinary" in the Cartan-Khaler theorem (the ranks of the restricted bundles corresponding to the Cartan characters). So, to be able to spot ordinary flag is to be able to integrate an EDS.

After this lengthy intro, I can finally formulate my

QUESTION: is there some algebraic machinery to construct $\mathrm{Fl}^\textrm{ord}(\mathcal{I})$ out of $\mathrm{Fl}(\mathcal{I})$? In particular, is there a test to recognise when the two are equal?

Some comments. Here's what I would expect: maybe there is some sort of "completion" $\mathcal{I}^\textrm{ord}\supseteq \mathcal{I}$, such that $$ \mathrm{Fl}^\textrm{ord}(\mathcal{I})=\mathrm{Fl}(\mathcal{I}^\textrm{ord})\, , $$ and I expect that $\mathcal{I}^\textrm{ord}$ can be costructed in a purely algebraic manner out of $\mathcal{I}$. Or, in a similar perspective, maybe there is a natural way to single out a "core" $K^\textrm{ord}(\mathcal{I})\subseteq K(\mathcal{I})$, such that $$ \mathrm{Fl}^\textrm{ord}(\mathcal{I})=K^\textrm{ord}(\mathcal{I})\cap\mathrm{Fl}(1,\ldots,n,V)\, . $$ In both cases, I have not even the slightest idea of how to construct these guys. I simply suspect that they exist.

The reason which led me to believe so, is that the "multi-bundled structure" of $\mathrm{Fl}(1,\ldots,n,V)$ is captured by the natural completely integrable distributions $\Delta_i$ it is equipped with (the maximal integral submanifolds of $\Delta_i$ being precisely the fibres of $(^*)$, for $i=1,\ldots,n$). So, in principle, it all boils down to study the integrability of the restricted distributions $\Delta_i|_{\mathrm{Fl}(\mathcal{I})}$, but I'm not able to go further.

Any reference to similar problems already dealt with in the literature, will be warmly appreciated!


1 Answer 1


The algebraic test for a flag to be ordinary that you are looking for is called Cartan's Test (also known as Cartan's Inequality). For example, see Chapter 3 of Exterior Differential Systems by Bryant, Chern, Gardner, Goldschmidt, and Griffiths.

As for whether one can construct an ideal $\mathcal{I}^{\mathrm{ord}}\subset\Lambda^*(V^*)$ whose integral flags are the ordinary integral flags of a given ideal $\mathcal{I}$, the answer is that, nearly always, one cannot. The ordinary integral flags of $\mathcal{I}$ are almost never a closed variety in the variety of all flags, but, for any ideal $\mathcal{J}\subset\Lambda^*(V^*)$, the variety of integral flags is always closed in the variety of all flags.

  • $\begingroup$ I asked this question after I finished reading Chap 3 of the book, being not satisfied by the Cartan's test, in this sense. Given a curve, I can decide whether a point belongs to its secant variety, by performing some test involving geometric manipulations; however, if you want to study the secant variety as a variety on its own, then an its algebraic description would be more desirable (even indispensable): to this end I can take, e.g., a parametrisation of the curve and obtain from it a parametrisation of the secant. I would like something like this for the variety of ordinary integral flag! $\endgroup$ Feb 24, 2016 at 14:56
  • $\begingroup$ Concerning your second remark, I agree with the fact that being an ordinary integral flag is an "open condition" (or, at least, "not closed"). However, I could rephrase my question, by asking how to construct the ideal of the complement, i.e., the (closed) variety of non-ordinary integral flags. (On a completely unrelated matter: Pawel really needs an answer from you). $\endgroup$ Feb 24, 2016 at 15:03
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    $\begingroup$ @GiovanniMoreno: I'll respond to both of your comments in this one comment: First, any condition one writes down that tests for a flag to be ordinary is going to be equivalent to Cartan's Test in some way or another. While many people have looked for ways to describe it in various terms using ideas from commutative algebra, in my opinion, none of these has been very useful in practice. Second, it can happen that the non-ordinary integral flags of $\mathcal{I}$, while closed in the variety of all flags, are not the integral flags of any ideal $\mathcal{J}$. $\endgroup$ Feb 24, 2016 at 15:50

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