Let $G$ and $H$ be two Hopf algebras, and $\pi: G \to H$ a Hopf algebra map. We will call an algebra of the form $$ M:= \lbrace m \in G ~ | ~ m_{(1)} \otimes \pi(m_{(2)}) = m \otimes 1 \rbrace $$ a quantum homogeneous space.

There are many well known examples of quantum homogeneous spaces where $G$ is a faithfully flat module over $M$: the quantum spheres, the quantum projective spaces, or more generally the quantum flag manifolds. What I would like are examples of quantum homogeneous spaces for which $G$ is not a faithfully flat module over $M$?

It is known that quantum-$SU(2)$ is not faithfully flat over some of the non-standard Podles spheres. However, these are not quantum homogenous spaces in the sense given above.

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    $\begingroup$ Do you require $\pi$ to be surjective? $\endgroup$
    – MTS
    Jul 22, 2012 at 1:38
  • $\begingroup$ No. I would even be interested in cases where $\pi$ is just a bialgebra map. $\endgroup$ Jul 22, 2012 at 1:50
  • $\begingroup$ @MTS: Despite the "G" and "H" notation, they are rings and not "geometric" objects, so the map $\pi$ is analogous in the algebraic geometry setting to the map between coordinate rings rather than their spectra. So rather than ask if $\pi$ is surjective, you probably meant to ask if it is assumed to be injective (for which the answer is probably "yes"...which the OP can confirm or not). $\endgroup$
    – user22479
    Jul 22, 2012 at 2:09
  • $\begingroup$ No, I meant to ask if the map was surjective. In that case one would regard $H$ as a "quantum subgroup" of $G$, and the algebra $M$ is a coideal subalgebra of $G$. If $G$ was the function algebra of a group, and $H$ the function algebra of the subgroup, $\pi$ would be the restriction map, and $M$ would be the subalgebra of functions invariant under translation by the subgroup, i.e. the function algebra of the corresponding homogeneous space; hence the term "quantum homogeneous space." $\endgroup$
    – MTS
    Jul 22, 2012 at 2:54
  • $\begingroup$ Interesting question - I guess I've only really thought about examples that were faithfully flat, so I don't have one off the top of my head. I will think about it. $\endgroup$
    – MTS
    Jul 22, 2012 at 3:19

1 Answer 1


Lets look at the commutative situation. So here is an example. I will change your notations -- otherwise I get too confused. Take $G=SL(N,\mathbb{C})$ and let $H$ be the subgroup of all those elements in $G$ for which the matrix entries strictly above the diagonal vanish and for which the diagonal entries are all $1$. The embedding $H\hookrightarrow G$ gives you a surjection $\pi:\mathbb{C}[G]\rightarrow \mathbb{C}[H]$ of Hopf algebras. The algebra $\mathbb{C}[G]$ has generators $a_{ij}$ for $1\le i,j\le N$ which are the coordinate functions of the matrix entries. The map $\pi$ maps $a_{ij}\mapsto 0$ if $i < j $ and $a_{ii}\mapsto 1$ for all $i$. All other generators $a_{ij}$ of $\mathbb{C}[G]$ are mapped to the corresponding generators of $\mathbb{C}[H]$. Your (quantum) homogeneous space is the subalgebra of $\mathbb{C}[G]$ generated by the last column, i.e. by the elements $a_{iN}$ for $1\le i\le N$. Lets call this subalgebra $\mathbb{C}[X]$. Then $\mathbb{C}[G]$ is not faithfully flat over $\mathbb{C}[X]$. Indeed, $\mathbb{C}[X]$ is just a polynomial ring in $N$ variables. The variables $a_{iN}$ generate a maximal ideal $\mathfrak{m}$ of $\mathbb{C}[X]$. Hence you have a proper inclusion $\mathfrak{m}\hookrightarrow \mathbb{C}[X]$. However, \begin{align} \mathfrak{m}\otimes_{\mathbb{C}[X]} \mathbb{C}[G] = \mathbb{C}[G] \end{align} because the tensor product on the left contains the determinant and hence the element $1$. This means that after tensoring over $\mathbb{C}[X]$ with $\mathbb{C}[G]$ your proper inclusion becomes an isomorphism. That contradicts faithfully flatness.

This example actually tells us more. The inclusion $\mathbb{C}[X]\hookrightarrow \mathbb{C}[G]$ gives us a map of the (maximal) spectra $G \rightarrow X$. If this map is not surjective (as in our example) then there exists a maximal ideal in $\mathbb{C}[X]$ which cannot be lifted to a maximal ideal in $\mathbb{C}[G]$. It's not hard to see, that in this case $\mathbb{C}[G]$ cannot be faithfully flat as a module over $\mathbb{C}[X]$.


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