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A Morse function is a map of a manifold to the real line locally equivalent to: $$f(x_1,\ldots, x_n)=-x_1^2\ldots -x_k^2+ x_{k+1}^2+\ldots+x_n^2$$ for some $k$. In other words, for which the singularities are as simple as possible. While a Lefschetz pencil is a map of a smooth projective variety to the projective line local analytically given by $f=x_1^2+\ldots+x_n^2$. So in this sense, they are very analogous. There are differences, however. Given a Morse function $f:X\to \mathbb{R}$, the collection of the above numbers $k$, called indices, determine the homotopy type*type.[the number of cells in a complex homotopic to $X$]. For "Artin's" vanishing, it is enough to choose an $f$, where these indices are bounded by dimension. (Details can be found in the first few pages of Milnor's Morse theory.) I'm not aware that there is any complete substitute on the other side, given say a Lefschetz pencil $f:X\to \mathbb{P}^1$. However, the pencil does give a way to calculate the (etale) cohomology of $X$ in terms of the critical points of $f$ and the monodromy, or more formally in terms of the direct images $R^if_*\mathbb{Q}_\ell$. And this is quite powerful.

*Added Note: this is a bit too imprecise. A bit more accurately, the list of indices of the critical points ordered by the values of $f$ determines the homotopy type.


Perhaps it would be more instructive give a Morse-like pseudo-proof of Artin's theorem. By "pseudo" I mean that there is step which I can't justify without a lot more effort than this is worth. Let $X\subset \mathbb{A}^n$ be an irreducible affine variety of dimension $n$ over an algebraically closed field. By generic projection, we get a nonconstant morphism $f:X\to \mathbb{A}^1$ which will play the role of our Morse function. Suppose that (*) the direct images $R^jf_*\mathbb{Q}_\ell$ were constructible and commuted with base change. Then by induction, $R^jf_*\mathbb{Q}_\ell=0$ for $j>n-1$. From the Leray spectral sequence, we need only prove that $H^i(\mathbb{A}^1,F)=0$ for $i>1$ and any constructible sheaf $F$, but this is easy. Note that if $f$ were proper, (*) would be automatic. In general, one could get around it by generic base change [SGA 41/2, p 236] and devissage.

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A Morse function is a map of a manifold to the real line locally equivalent to: $$f(x_1,\ldots, x_n)=-x_1^2\ldots -x_k^2+ x_{k+1}^2+\ldots+x_n^2$$ for some $k$. In other words, for which the singularities are as simple as possible. While a Lefschetz pencil is a map of a smooth projective variety to the projective line local analytically given by $f=x_1^2+\ldots+x_n^2$. So in this sense, they are very analogous. There are differences, however. Given a Morse function $f:X\to \mathbb{R}$, the collection of the above numbers $k$, called indices, determine the homotopy typetype*. For "Artin's" vanishing, it is enough to choose an $f$, where these indices are bounded by dimension. (Details can be found in the first few pages of Milnor's Morse theory.) I'm not aware that there is any complete substitute on the other side, given say a Lefschetz pencil $f:X\to \mathbb{P}^1$. However, the pencil does give a way to calculate the (etale) cohomology of $X$ in terms of the critical points of $f$ and the monodromy, or more formally in terms of the direct images $R^if_*\mathbb{Q}_\ell$. And this is quite powerful.

*Added Note: this is a bit too imprecise. A bit more accurately, the list of indices of the critical points ordered by the values of $f$ determines the homotopy type.


Perhaps it would be more instructive give a Morse-like pseudo-proof of Artin's theorem. By "pseudo" I mean that there is step which I can't justify without a lot more effort than this is worth. Let $X\subset \mathbb{A}^n$ be an irreducible affine variety of dimension $n$ over an algebraically closed field. By generic projection, we get a nonconstant morphism $f:X\to \mathbb{A}^1$ which will play the role of our Morse function. Suppose that (*) the direct images $R^jf_*\mathbb{Q}_\ell$ were constructible and commuted with base change. Then by induction, $R^jf_*\mathbb{Q}_\ell=0$ for $j>n-1$. From the Leray spectral sequence, we need only prove that $H^i(\mathbb{A}^1,F)=0$ for $i>1$ and any constructible sheaf $F$, but this is easy. Note that if $f$ were proper, (*) would be automatic. In general, one could get around it by generic base change [SGA 41/2, p 236] and devissage.

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A Morse function is a map of a manifold to the real line locally equivalent to: $$f(x_1,\ldots, x_n)=-x_1^2\ldots -x_k^2+ x_{k+1}^2+\ldots+x_n^2$$ for some $k$. In other words, for which the singularities are as simple as possible. While a Lefschetz pencil is a map of a smooth projective variety to the projective line local analytically given by $f=x_1^2+\ldots+x_n^2$. So in this sense, they are very analogous. There are differences, however. Given a Morse function $f:X\to \mathbb{R}$, the collection of the above numbers $k$, called indices, determine the homotopy type. For "Artin's" vanishing, it is enough to choose an $f$, where these indices are bounded by dimension. (Details can be found in the first few pages of Milnor's Morse theory.) I'm not aware that there is any complete substitute on the other side, given say a Lefschetz pencil $f:X\to \mathbb{P}^1$. However, the pencil does give a way to calculate the (etale) cohomology of $X$ in terms of the critical points of $f$ and the monodromy, or more formally in terms of the direct images $R^if_*\mathbb{Q}_\ell$. And this is quite powerful.


Perhaps it would be more instructive give a Morse-like pseudo-proof of Artin's theorem. By "pseudo" I mean that there is step which I can't justify without a lot more effort than this is worth. Let $X\subset \mathbb{A}^n$ be an irreducible affine variety of dimension $n$ over an algebraically closed field. By generic projection, we get a nonconstant morphism $f:X\to \mathbb{A}^1$ which will play the role of our Morse function. Suppose that (*) the direct images $R^jf_*\mathbb{Q}_\ell$ were constructible and commuted with base change. Then by induction, $R^jf_*\mathbb{Q}_\ell=0$ for $j>n-1$. Then from From the Leray spectral sequence, we need only prove that $H^i(\mathbb{A}^1,F)=0$ for $i>1$ and any constructible sheaf $F$, but this is easy. Note that if $f$ were proper, (*) would be automatic. In general, one could get around it by generic base change [SGA 41/2, p 236] and devissage.

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