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I suppose I am the first one who asked about Weil representation here.

In studying Weil representation, I fell into a slough and so determined to ask you for shedding a light. I think your responses would be a gleam of hope for many others struggling exactly at the same part like me.

Let me recall some basic notations required to set-up the Weil representation.

Let F be a p-adic field and E its quadratic extension field. Let V be a n-dimension hermitian vector space over E and W m-dimension skew-hermitian vector space over E and U(V) and U(W) are their isometry groups respectively.

Let $\mathbb{W}=Res_{E/F} (V \otimes W)$ along with a complete polarization $\mathbb{W}=\mathbb{X} \oplus \mathbb{Y}$.

Let $Sp(\mathbb{W}) $be the symplectic group of $\mathbb{W}$ and $\tilde{Sp}(\mathbb{W})$ be its two-fold metaplectic cover.

Then, $\tilde{Sp}(\mathbb{W})$ has the Weil representation on $\mathbb{S}(\mathbb{X})$, the Schwartz space on $\mathbb{X}$, and its law of action is well known explicitly.

When m is even, we can find easily the complete polarization of $\mathbb{W}=\mathbb{X} \oplus \mathbb{Y}$ as follows;

If we let $W=X \oplus Y$, the complete polarization of W, then $\mathbb{X}=V \otimes X, \mathbb{Y}=V \otimes Y$ gives the above polarization of $\mathbb{W}$.

My question arises here.

If m is not even but odd, how can we find the polarization of $\mathbb{W}$? And in this case, the explicit Weil representation law is known?

I searched a lot for this, but found no paper which dealt this.

If you know the technics to manage this case or some paper treatng this case, would you let me know it? Then I will be very grateful for your benovelence and it may also helps many others.

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I am a little confused about your explanation of the case where $m$ is even. Don't you have to take a restriction of scalars before getting $\mathbb{W}$? – S. Carnahan Dec 12 '12 at 9:40
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I do not fully understand your construction of $\mathbb W$, but it seems to be a symplectic space. For such spaces polarizations always exist. Also, you do not need a polarization to define the Weil representations (see Tata Lectures on Theta III, by Mumford). – Amritanshu Prasad Dec 13 '12 at 4:09
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"and E its quadratic extension field" : local fields don't have a unique quadratic extension (unless you are over the reals). – Chandan Singh Dalawat Dec 16 '12 at 2:21
up vote 1 down vote accepted

At least as a place-holder answer: the issues may be more about just the geometric algebra rather than the Segal-Shale-Weil/oscillator repn. We can multiply hermitian and skew-hermitian forms by a non-zero element flipped in sign by Galois, making hermitian into skew and vice-versa, so there is no distinction between the two.

It is necessary to understand the non-degenerate skew-symmetric form on the restricted-scalars tensor product. To my mind, the story is clearest if we tensor two hermitian spaces, and take the "imaginary part" of the hermitian form on the tensor product: the restricted scalars $H={\rm Res}(V\otimes_E W)$ has alternating form $\langle v\otimes w,v'\otimes w'\rangle={\rm im}\big(\langle v,v'\rangle\cdot \langle w,w'\rangle\big)$, where ${\rm im}\alpha=(\alpha-\alpha^\sigma)/2\sqrt{D}$, with $\sigma$ the non-trivial Galois element and $E=F(\sqrt{D})$.

Thus, for any expressions of $V,W$ as extensions of scalars from $F$-vectorspaces ("realforms") $V_o,W_o$ of the same dimension, with the hermitian forms obtained naturally by extending scalars "sesquilinearly", the image in $H$ of $E\cdot (V_o\otimes_F W_o)$ is a totally isotropic subspace of $H$, etc.

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Dear Garret, Thanks very much for sheding a light. I think your answer touches the point. – anonymous Dec 14 '12 at 4:12

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