integral basis for the Lie algebra of the Neron model of an abelian variety

Let $A$ be an abelian variety over a number field $K$. Let $\mathcal{A}$ be the Neron model of $A$ over $O_K$. Let $\Omega_{\mathcal{A}/O_K}$ be the sheaf of invariant differential forms on $\mathcal{A}$.

In the formulation of the Tamagawa Number Conjecture for abelian varieties, for e.g., page 13 of the following paper http://arxiv.org/abs/math/0507275, we have to take an integral basis for the Lie algebra $Lie(\mathcal{A})$ in order to compute the period, which is the determinant of the isomorphism $\alpha_{A} : H_B^+(A)_{\mathbb{R}} \to Lie(A)_{\mathbb{R}}$.

In that paper, where $K=\mathbb{Q}$, the author just takes a $\mathbb{Z}$-basis of $Hom_{\mathbb{Z}}(\Omega_{\mathcal{A}/O_K}(\mathcal{A}),\mathbb{Z})$ as the integral basis for $Lie(A)$. I guess you can do this because $Hom_{\mathbb{Z}}(\Omega_{\mathcal{A}/O_K}(\mathcal{A}),\mathbb{Z})$ is a free $\mathbb{Z}$-module.

My question is that in general when $K\neq \mathbb{Q}$ and $\Omega_{\mathcal{A}/O_K}(\mathcal{A})$ is not a free $O_K$-module but only a projective $O_K$-module, how can we take an integral basis for $Lie(A)$ ?

Thank you very much.

The same problem already appears in the formulation of the Birch and Swinnerton-Dyer conjecture for say an elliptic curve over a number field. When there is no longer a global minimal Weierstrass model, then we do not have a invariant differential $\omega_E$ that generates the $O_K$-module of differentials on the Néron model of $E$.
One way to formulate these conjectures is to choose a $K$-basis of the differentials formed of integral differentials. The dual basis $\delta$ (in the notation of Venjakob's survey) is a $O_K$-lattice in Lie${}_{O_K}(A^{\vee})$, which becomes a $K$-basis after $\otimes K$.
Now one can integrate paths on $A(\mathbb{C})$ with respect to these differentials to get periods. The same choice of $\delta$ will appear later in the $\mathfrak{p}$-adic version for finite places $\mathfrak{p}$. This is (3.30) in the linked survey. The local determinants there are now only equal to the Tamagawa number (defined as the number of components of the Néron model at $\mathfrak{p}$) if the basis happens to be an $O_{\mathfrak{p}}$-basis of the differentials on the Néron model. Otherwise, there will be a correction factor taking this into account.
For instance for an elliptic curve, one can fix any integral invariant differential $\omega$ (i.e. fix any integral Weierstrass equation). Define the real and complex periods with respect to $\omega$. Divide the leading term of the $L$-function by the appropriate product of periods. The conjecture now says that this is equal to the usual formula, excet that the Tamagawa numbers at places where the equation was not minimal have to be modified. See for instance page 5 of http://arxiv.org/abs/math/0610290.
• Thank you very much for your answer. Does that mean, there is no integral basis for $Lie(A)$ in the general case then ? Mar 11, 2015 at 3:18
• Lie$(A^{\vee})$ is defined to be the $O_K$-dual of the differentials on the Néron model. So if that is not free, neither will Lie$(A^{\vee})$. But a basis of Lie${}_K(A^{\vee})$ is all that you need. Mar 11, 2015 at 10:18