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If $X$ is an affine scheme over the field $k$ than algebraic invariants of the coordinate ring $k[X]$ usually have a geometric interpretation in terms of $X$ (and vice versa). As an example, the minimal primes of $k[X]$ correspond to the irreducible components of $X$.

Now suppose $G$ is a finite group scheme over $k$. Thus $k[G]$ is a finite dimensional Hopf algebra and by a well-known theorem of Larson-Sweedler, $k[G]$ has a non-zero integral, i.e. an element $a_0 \in k[G], a_0 \neq 0$ such that $a\cdot a_0 = \epsilon(a)a_0$ for all $a \in k[G]$, where $\epsilon: k[G] \to k$ is the augementation induced by the identity $e: \operatorname{Spec}(k) \to G$.

Is there a colorful geometric interpretation of this integral ?

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You can always take $a_0=0$, so it looks like you forget a condition (maybe just that $a_0$ is nonzero). – JBorger Jul 4 '11 at 22:22
That's it. Thanks, James. – Ralph Jul 4 '11 at 22:27
Geometrically, a finite group scheme realizes over any field extension of $k$ as a finite set of points. As far as I understand, the integrals on $k\left[G\right]$ corresponds to functions on this set that are zero everywhere except of the unity of the group. Hence, the existence of a nonzero integral is more or less the fact that the indicator function of the unity of a finite group can be realized as a polynomial function. But this is probably not particularly geometric, or I wonder why nobody else has posted this in two months... – darij grinberg Sep 3 '11 at 13:00
@darijgrinberg: I would say your comment is accurate for etale group schemes, but rather more subtle for general finite group schemes. Though maybe geometrically non-reduced group schemes are not so "geometric" anyway... – Daniel Litt Sep 1 '13 at 3:56

Just a quick answer: I have maybe slightly different Hopf algebras in mind as you, but in my applications the integral often behaves like the fundamental class of a manifold.

[Added as answer:] The example I have in mind is as follows: Take $H=\mathcal{U}_q(\mathfrak{g})$, for $q$ a root of unity the finite-dimensional quotient (Frobenius-Lusztig kernel) $\bar{H}$ and $\mathfrak{B}(M)$ the Borel part, the so-called Nichols algebra (generated be all $E_i$'s). The integral of $\mathfrak{B}(M)$ corresponds to the longest element in the Weyl group. I have read people connect that to Schubert cells but I can not repeat these arguments.

You can use it to prove e.g. Poincare Duality of $\mathfrak{B}(M)$. The Hilbert series hence exhibit the nice palindrom symmetry.

I'm very sure you can argue similarly for non-truncations, but then it seems to me you have to do topological completion for that. Maybe the regular functions on a Lie group give you a more accessible example? Generally it should be a good idea to think of the integral in terms of a Frobenius algebra!

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How are your Hopf algebras related to a manifold ? – Ralph Oct 31 '13 at 21:58
added to answer – Simon Lentner Nov 4 '13 at 10:13

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