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Let $E$ be an elliptic curve over $\mathbf{Q}$ which has split multiplicative reduction at $p$ (a prime). If one chooses a global Neron model of $E$ over $\mathbf{Z}$ (unique up to unique isomorphism over $\mathbf{Z}$) then one gets a holomorphic differential (on the Neron model) which is well defined up to $\pm 1$. Then one may define the following positive real number $$ \Omega_E:=\int_{E(\mathbf{R})}|\omega| \in \mathbf{R}_{>0}. $$

The quantity $\Omega_E$ "seems to appear" in the $p$-adic formulation of BSD. Let $\alpha,\alpha'\in\mathbf{C}$ be the two roots of the Frobenius at $p$ (acting on the Tate module). Let us assume that $\alpha$ is ordinary i.e. that its image in $\mathbf{C}_p$ is a unit ($v_p(\alpha)=0$). Note here that the choice $\alpha$ (the ordinary root at $p$) depends (in general) on the choice of an embedding $\bar{\mathbf{Q}}\rightarrow \mathbf{C}_p$. Once such an embedding is fixed then the $p$-adic $L$-function $L_p(E,s)$ (which is defined by $p$-adic integration using a certain measure which depends on $\alpha$), seems to be, as far as I understand, completely determined (may be this is here that I make a mistake). Therefore "the image" of $\Omega_E$ in $\mathbf{C}_p$ should also be completely determined. But if $\Omega_E$ is transcendental over $\mathbf{Q}$ then one can send $\Omega_E$ on any transcendental (over $\mathbf{Q}$) element of $\mathbf{C}_p$ by choosing a suitable embedding of $\mathbf{R}\hookrightarrow \mathbf{C}_p$.

Q: So how can we reconcile the fact that $\Omega_E$ admits many different embeddings inscide $\mathbf{C}_p$ with the observation that its image in $\mathbf{C}_p$ should be completely determined?

added: Is this discrepancy somehow compensated by the presence of the $\mathcal{L}$ invariant...?

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up vote 4 down vote accepted

Mazur-Tate-Teitelbaum have written their p-adic BSD paper for a modular form of even weight $k\geq 2$ so unless we are dealing with a elliptic curve, we might want to avoid choosing a period. They view modular symbols are taking values in $L = L_f\otimes \mathbb{Q}$, which is a $2$-dimensional $\mathbb{Q}$-vector space in $\mathbb{C}$ with $L_f$ being the lattice of values of the modular symbol. As the $p$-adic measure is constructed from linear combinations of modular symbols with coefficients using $\alpha\in\overline{\mathbb{Q}_p}$, it takes values in $L\otimes\mathbb{C}_p$. See top of page 13.

Now if $E$ is an elliptic curve, we have a canonical choice of a basis of $L$ given by the real and imaginary Néron periods. Hence the values of the $p$-adic $L$-function can be written as a $\mathbb{C}_p$-linear combination of these two periods. Maybe the confusion would not arise if they wrote the $p$-adic BSD for $E$ on page 38 as $L^{(r)}_p(E) = \text{some $p$-adic number} \otimes \Omega_E^{+} $ inside $\mathbb{C}_p\otimes L$.

In their choice, the $p$-adic $L$-function only depends on the modular form $f$, and hence would not change by an isogeny $E\to E'$. An other normalisation, the one I use more often, would be to divide the modular symbols by the periods from the beginning to obtain rational numbers $[r]^{+}$ and $[r]^-$ attached to a fixed elliptic curve $E$. Then the period would not appear in the $p$-adic BSD formula anymore, but the $p$-adic $L$-function is then depending on the curve in the isogeny class. From the perspective of Iwasawa theory that may be better, because this version is the characteristic series of some Selmer group.

So I don't think your question relates to the $\mathcal{L}$-invariant.

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Thanks Chris for your nice explanation which just reconciled my mind with the p-adic BSD! I can see what you win with your normalization (less confusion), but on the other hand I very much like the isogeny invariance. Unfortunately, one has to choose... – Hugo Chapdelaine Aug 1 '13 at 23:54

You should make precise what version of p-adic BSD you are looking at. If you mean e.g. the one from Mazur-Tate-Teitelbaum of the form $L'_p(1) = \mathcal{L}_pL_{\infty}(1)/\Omega$, I guess the point is that, the usual BSD predicts that $L_{\infty}(1)/\Omega$ is a rational number, so that has a natural p-adic interpretation and you don't need to interpret the numerator and denominator as p-adic numbers separately.

Stupid TeX is not formatting right...

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I see. So in my question I was referring to the Mazur-Tate-Teitelbaum paper. It is on p. 38. So I guess that your definition of $\mathcal{L}_p$ is not the same as the one used in their paper, otherwise the formulas don't match. – Hugo Chapdelaine Aug 1 '13 at 17:48
I just looked up the paper of Greenberg and Stevens. Maybe the original statement in MTT is different. – Felipe Voloch Aug 1 '13 at 18:28

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