Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T and let's live in the category of commutative algebraic groups over k.

Out of four expressions like [Gm => [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).

To prove equality (*) use for B = Gm

   (**)              A \otimes [B=>B] ==== [[A=>B] => B].

Question: Is there an example of commutative algebraic group B, other than Gm, for which the identity (**) is also true or perhaps true in some other sense?

(One thing I specifically have in mind is that if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, as if it was a with any rigid tensor category, the formula would hold for all A and B)

Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T and let's live in the category of commutative algebraic groups over k.

Out of four expressions like [Gm => [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).

The equality (*) can be proven by using the following formula with B = Gm

   (**)              A \otimes [B=>B] ==== [[A=>B] => B].

Question: Is there another example of commutative algebraic group or a similar generalized object B, for which the identity (**) is true or true in some generalized sense?

One thing I specifically have in mind is that if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, as if they were vector spaces, the formula would hold for all A and B. So, what's a category that is related to algebraic groups but which posesses this property?

Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T and let's live in the category of commutative algebraic groups over k.

Out of four expressions like [Gm => [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).

To prove equality (*) use for B = Gm

   (**)              A \otimes [B=>B] ==== [[A=>B] => B].

Question: Is there an example of commutative algebraic group B, other than Gm, for which the identity (**) is also true or perhaps true in some other sense?

(One thing I specifically have in mind is that if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, as if it was a with any rigid tensor category, the formula would hold for all A and B)

Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T and let's live in the category of commutative algebraic groups over k.

Out of four expressions like [Gm => [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).

To prove equality (*) use for B = Gm

   (**)              A \otimes [B=>B] ==== [[A=>B] => B].

Question: Is there an example of commutative algebraic group B, other than Gm, for which the identity (**) is also true or perhaps true in some other sense?

(One thing I specifically have in mind is that if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, as if it was a with any rigid tensor category, the formula would hold for all A and B)

This question comes from my notes, thus slightly unusual structure.

Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T.

Out of four expressions like [Gm => [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).

To prove equality (*) use

   (**)       A \otimes [B=>B] ==== [[A=>B] => B]

for B = Gm.

Question: Is there an example of algebraic group B, other than Gm, for which the identity (**) is also true (or perhaps true in some other sense)?

(Note that would be true if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, in analogy with vector spaces, but I don't think you can do this to groups)

Let's use the notation of [A=>B] for Hom(A, B). Take a 1-dimensional algebraic torus Gm and higher-dimensional torus T and let's live in the category of commutative algebraic groups over k.

Out of four expressions like [Gm => [Gm=>T]] etc. half give back T (*), others the dual torus TV, in the sense that X*(T) := [Gm => T] = [TV => Gm] =: X*(TV).

To prove equality (*) use for B = Gm

   (**)              A \otimes [B=>B] ==== [[A=>B] => B].

Question: Is there an example of commutative algebraic group B, other than Gm, for which the identity (**) is also true or perhaps true in some other sense?

(One thing I specifically have in mind is that if we could write [X => Y] = X* \otimes Y whenever X and Y are groups, as if it was a with any rigid tensor category, the formula would hold for all A and B)