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This was meant to be an innocenta comment, but it won't fit, so here we go.

I'd just like to humbly point outJust a few naive observations.

Let's reorganize thoughts as followsFor completeness:

This was meant to be an innocent comment, but it won't fit, so here we go.

I'd just like to humbly point out a few naive observations.

Let's reorganize thoughts as follows:

This was meant to be a comment, but it won't fit, so here we go.

Just a few naive observations.

For completeness:

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EDIT: I should also add that several of the averages (both unconditional and conjectural) that are key to the Bhargava-Skinner-Zhang methods, fail in char $p$, while I'd regard a method towards BSD to be "promising", if it were able to make progress or settle its char $p$ counterpart, first.

EDIT: I should also add that several of the averages (both unconditional and conjectural) that are key to the Bhargava-Skinner-Zhang methods, fail in char $p$, while I'd regard a method towards BSD to be "promising", if it were able to make progress or settle its char $p$ counterpart, first.

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Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional. Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.

Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.

Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional. Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.

Regardless, clearly Super-weak BSD, which is what the discussion in the question, comments, and around the Bhargava-Skinner-Zhang paper, were about, is not equivalent to weak BSD under any circumstances, no matter that $K = \mathbf{Q}$ and $A$ is $1$-dimensional.

Showing 100% of elliptic curves over $\mathbf{Q}$ satisfy the BSD conjecture does not show weak BSD (as the OP took care to make clear in his/her question). Such result, if ever available, should be regarded as being motivational only.

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