In his famous article [1] Klein constructs a representation of $G=PSL_2(\mathbb{F}_7)$ in $\mathbb{C}^3$ (of which the first invariant polynomial of three variables gives rise to the famous Klein's quartic).

All other irreducible representations of $G$ are very simple or natural to obtain: the action on $\mathbb{P}^1_{\mathbb{F}_7}$ produces a representation of dimension $7$. The isomorphism $G\simeq GL_3(\mathbb{F}_2) = PGL_3(\mathbb{F}_2)$ and the action on the Fano plane $\mathbb{P}^2_{\mathbb{F}_2}$ produces a representation of dimension $6$. There's the trivial representation, one of dimension $8$ that is simply induced from the normalizer of a $7$-Sylow subgroup (which can be also thought geometrically as these subgroups are point-stabilizers of the above mentioned actions). The two missing are Klein's representation and its conjugate. However, to construct those we appeal to explicit generators and relations by writing explicit $3\times 3$ matrices that correspond to the generators of $G$ of orders $7,3$ and $2$ (this last involution is the hard one to find).

Is there a geometric intuition behind this representation? and if not (as it seems to be) is there an inherent reason for this representation being "hard" to obtain?

In the fantastic article Elkies explains this last involution as the discreet Fourier transform on the space of functions $\mathbb{F}_7 \rightarrow \mathbb{C}$. To my mind this appearance of a Fourier transform is indication of some deeper explanation of this action involving some more elaborate geometric tricks as Mukai transforms more than an explanation itself. Many things that you may want to know about $G$ or Klein's quartic are beautifully explained in that article of Elkies.

Finally, it is worth noticing that this phenomenon occurs for other groups of the form $PSL_2(\mathbb{F}_q)$ where most irreducible representations are geometric in nature, except the few ones that you need to write by hand the generators.

[1]: F. Klein, “Ueber die Transformationen siebenter Ordnung der elliptischen Funktionen”, Math. Annalen 14 (1879), 428–471.

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    $\begingroup$ If you look at weight two cusp forms for the principal congruence subgroup $\Gamma(7)$ of $SL_2(\mathbb Z)$, they form a dimension three representation. This seems geometric enough to me. $\endgroup$ Nov 5, 2014 at 12:35
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    $\begingroup$ Klein's representations seem to be the cuspidal representations of G (in the sense that they are not subrepresentations of parabolically induced representations). Parabolically induced representations have simple geometric models for they may be considered as spaces of functions on flag varieties. On the other hand, the cuspidal representations are the "difficult ones". They do not have simple geometric models. They are obtained in the $l$-adic cohomology of Deligne-Lusztig varieties. $\endgroup$ Nov 5, 2014 at 12:49
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    $\begingroup$ Perhaps section 11.4.4 in "Representations of $SL_2(\mathbb{F}_q)$" by Cédric Bonnafé is interesting to you. He identifies $PSL_2(\mathbb{F}_7) \times \mathbb{Z}/2\mathbb{Z}$ as an exceptional complex reflection group of rank 3 ($G_{24}$ on the Shephard-Todd list). $\endgroup$ Nov 5, 2014 at 13:10
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    $\begingroup$ These representations arise as a constituent of the Weil representation of $\operatorname{SL}(2,q)$ ($q$ odd). The Weil rep. decomposes into two irreps of dims $(q\pm 1)/2$, where one irrep has $-I$ in the kernel and can thus be viewed as irrep of $\operatorname{PSL}(2,q)$. And the Weil rep. sends elements of order $4$ in $\operatorname{SL}(2,q)$ to the Fourier transform. See arxiv.org/abs/0903.1486 for some elementary proofs about the Weil rep in the finite case. $\endgroup$ Nov 5, 2014 at 13:57
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    $\begingroup$ Just a tiny footnote to Frieder Ladisch's comment: whichever one of $\frac{q \pm 1}{2}$ is odd is the dimension of the irreducible representation of ${\rm PSL}(2,q)$, so when $q = 7$ that dimension is $3.$ The Weil representation of ${\rm SL}(2,q)$ has dimension $q$ and its two non-trivial irreducible constituents are the representations on the eigenspaces of the central involution. $\endgroup$ Nov 5, 2014 at 14:25

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I guess the answer depends on what you call "geometric"... If you accept some basic algebraic geometry, you can do the following. Consider the homographs $\ \alpha :z\mapsto z+1\ $ and $\ \beta : z\mapsto -1/z\ $ of $\ \mathbb{P}^1_{\mathbb{F}_7}$. We have $\alpha ^7=\beta ^2=(\alpha \beta )^3=1$; it is an easy exercise to show that $\alpha $ and $\beta $ generate $PSL_2(\mathbb{F}_7)$. Thus we have a surjective map $\pi _1(\mathbb{P}^1\smallsetminus\{0,1,\infty\} )\rightarrow PSL_2(\mathbb{F}_7)$, hence a Galois covering $C\rightarrow \mathbb{P}^1$ with group $PSL_2(\mathbb{F}_7)$, branched along $\{0,1,\infty\}$. The Riemann-Hurwitz formula gives $g(C)=3$. The group $PSL_2(\mathbb{F}_7)$ acts on $C$, hence on the space of holomorphic forms $\Omega _C$, and this is the Klein representation.

  • $\begingroup$ Thanks! this is is indeed a nice answer. Lev's comment on cusps forms above is also nice in that it interprets the appearance of the Fourier transform. $\endgroup$ Nov 6, 2014 at 9:35

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