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Given some non-empty set $X$, does the set of positive definite kernels on $K_X$ form a ring with pointwise addition and multiplication. I am convinced it does not as surely if $k \in K_X$ then we would require that $-k \in K_X$ also, which I don't think is true?

I ask because Hagan (the user) seems to think it does in this question: The Polynomial Kernel

If not, does it even form an abelian group with respect to any other operation? Or better yet, a ring?

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    $\begingroup$ It isn't a ring: Hagen's statement was not quite correct. It is closed under pointwise addition, multiplication by positive reals, and pointwise product. $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2019 at 14:41
  • $\begingroup$ I think that this question, while natural, should really have been a comment on the old question (but I appreciate that as a new user you can't yet leave comments) $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2019 at 14:41
  • $\begingroup$ Thank you, yes I had tried to leave a comment but was unable. $\endgroup$
    – Jack O'Connor
    Commented Apr 20, 2019 at 16:57
  • $\begingroup$ It's an abelian group for every abelian group structure... every nonempty set carries an abelian group structure (and even a ring structure). So "any other operation" is too open-ended, since it's not really interesting if the ring structure is unrelated to the way this set was defined $\endgroup$
    – YCor
    Commented Apr 20, 2019 at 23:50

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To make it answered (cw):

It's not stable under taking $x\mapsto -x$ so is not a ring, as it's not a group under addition.

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