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Georges Elencwajg
Georges Elencwajg
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Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice.

Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even $\ldots$... [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.]

So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what.

Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is $w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$.

This leads me to ask the question:

If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible?

Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice.

Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even $\ldots$ [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.]

So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what.

Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is $w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$.

This leads me to ask the question:

If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible?

Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice.

Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even ... [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.]

So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what.

Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is $w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$.

This leads me to ask the question:

If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible?

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Georges Elencwajg
Georges Elencwajg
  • 47.5k
  • 14
  • 159
  • 241

Are the Stiefel-Whitney classes of a vector bundle the only obstructions to its being invertible?

Consider vector bundles on connected paracompact topological spaces. Such a vector bundle $E$ on $X$ is said to be invertible if there exists some other bundle $F$ whose sum with $E$ is trivial: $E\oplus F \simeq \epsilon ^N $. The terminology "invertible" (used by Tammo tom Dieck for example) comes from K-theory and is not so weird as it looks:in $\tilde K(X)$ the class of $F$ is indeed the additive inverse of that of $E$. If all vector bundles on $X$ are invertible, then every class (=virtual bundle) in $\tilde K(X)$ is represented by an actual bundle, which is rather nice.

Now, every vector bundle is invertible if $X$ is compact or is a differentiable manifold or even a topological manifold or even a subspace of some $\mathbb R^n$ or even a space of finite combinatorial Lebesgue dimension or even $\ldots$ [Please correct me if I'm wrong: this is an interpretation/synthesis of what I read, sometimes between the lines, in several places.]

So one might optimistically hope that every vector bundle on a paracompact connected space is invertible: after all, what could go wrong? Here is what.

Consider $X=\mathbb {RP}^{\infty}$ (infinite dimensional real projective space) and the tautological line bundle $\gamma$ on $X$. Its total Stiefel-Whitney class is $w(\gamma)=1+x \in H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$, where $x$ is the first Stiefel-Whitney class of $\gamma$ . If $\gamma$ had an inverse vector bundle $F$ we would have $w(\gamma) w(F)=1$ and this is impossibl since $w(\gamma)=1+x$ is not invertible in the cohomology ring $H^\ast (\mathbb {RP}^{\infty},\mathbb Z /2)=(\mathbb Z /2)[x]$ ( a polynomial ring in one indeterminate over $\mathbb Z /2)$.

This leads me to ask the question:

If a vector bundle on a connected paracompact space has a total Stiefel-Whitney class invertible in its cohomology ring, does it follow that the bundle itself is invertible?