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If you're talking about Grassmannians, then you can use the direct sum maps

$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$

to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.

According to Symmetric polynoms are Hopf algebra ? What for one needs co-product ?Symmetric polynoms are Hopf algebra ? What for one needs co-product ? the bialgebra is enough to get the whole Hopf structure.

Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).

If you're talking about Grassmannians, then you can use the direct sum maps

$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$

to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.

According to Symmetric polynoms are Hopf algebra ? What for one needs co-product ? the bialgebra is enough to get the whole Hopf structure.

Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).

If you're talking about Grassmannians, then you can use the direct sum maps

$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$

to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.

According to Symmetric polynoms are Hopf algebra ? What for one needs co-product ? the bialgebra is enough to get the whole Hopf structure.

Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).

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Steven Sam
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If you're talking about Grassmannians, then you can use the direct sum maps

$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$

to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.

According to Symmetric polynoms are Hopf algebra ? What for one needs co-product ? the bialgebra is enough to get the whole Hopf structure.

Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).