Skip to main content
added 25 characters in body
Source Link
Peter Crooks
  • 4.9k
  • 2
  • 22
  • 42

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general (ie. for positive genus), this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general (ie. for positive genus), this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

added 4 characters in body
Source Link
Peter Crooks
  • 4.9k
  • 2
  • 22
  • 42

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

added 24 characters in body
Source Link
Peter Crooks
  • 4.9k
  • 2
  • 22
  • 42

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore consists ofcontains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

There are holomorphic line bundles over a compact Riemann surface $X$ that are not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore consists of smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.

One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general, this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

Source Link
Peter Crooks
  • 4.9k
  • 2
  • 22
  • 42
Loading