Karl, my feeling is that this will not be true without further conditions. Here are some thoughts:

1. The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true.

2. I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$.

3. So, let's see how one could construct a counterexample. Perhaps cones will do...

4. Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all $0<i<d-1$ and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM.

5. So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_*\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism.

6. So what condition should we ask for? I guess the first guess is something like $R^i\pi_*\mathcal O_Y=0$ for $i>0$. I think this might give you what you want: Grothendieck duality gives that then $$R\pi_*\omega_Y^\cdot\simeq_{qis}\omega_X^\cdot$$ Now if $X$ is CM, then the right hand side has only one non-zero cohomology. Now one could write $\omega_Y^\cdot$ as $$\omega_Y[n]\to \omega_Y^\cdot \to \omega_Y^+ \to^{+1} $$ So, if we also knew that $R^i\pi_*\omega_Y=0$ for $i>0$, then it would follow that $R\pi_*\omega_Y^+=0$ and I think that should imply that actually $\omega_Y^+=0$ which would imply that $Y$ is CM. So, this seems like a condition:

If $R^i\pi_*\mathcal O_Y=0$ and $R^i\pi_*\omega_Y=0$ for $i>0$, then what you want might follow. Then again, this might be more then what you would want to assume.

Any thoughts?

Karl, my feeling is that this will not be true without further conditions. Here are some thoughts:

1. The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true.

2. I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$.

3. So, let's see how one could construct a counterexample. Perhaps cones will do...

4. Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all $0<i<d-1$ and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM.

5. So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_*\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism.

6. So what condition should we ask for? I guess the first guess is something like $R^i\pi_*\mathcal O_Y=0$ for $i>0$. I think this might give you what you want: Grothendieck duality gives that then $$R\pi_*\omega_Y^\cdot\simeq_{qis}\omega_X^\cdot$$ Now if $X$ is CM, then the right hand side has only one non-zero cohomology. Now one could write $\omega_Y^\cdot$ as $$\omega_Y[n]\to \omega_Y^\cdot \to \omega_Y^+ \to^{+1} $$ So, if we also knew that $R^i\pi_*\omega_Y=0$ for $i>0$, then it would follow that $R\pi_*\omega_Y^+=0$ and I think that should imply that actually $\omega_Y^+=0$ which would imply that $Y$ is CM. So, this seems like a condition:

If $R^i\pi_*\mathcal O_Y=0$ and $R^i\pi_*\omega_Y=0$ for $i>0$, then what you want might follow. Then again, this might be more then what you would want to assume.

Any thoughts?

Karl, my feeling is that this will not be true without further conditions. Here are some thoughts:

1. The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true.

2. I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$.

3. So, let's see how one could construct a counterexample. Perhaps cones will do...

4. Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all $0<i<d-1$ and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM.

5. So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_*\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism.

6. So what condition should we ask for? I guess the first guess is something like $R^i\pi_*\mathcal O_Y=0$ for $i>0$. This still does not make it clear, but at least rules out the above example.

Any thoughts?

Karl, my feeling is that this will not be true without further conditions. Here are some thoughts:

1. The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true.

2. I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$.

3. So, let's see how one could construct a counterexample. Perhaps cones will do...

4. Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all $0<i<d-1$ and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM.

5. So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_*\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism.

6. So what condition should we ask for? I guess the first guess is something like $R^i\pi_*\mathcal O_Y=0$ for $i>0$. I think this might give you what you want: Grothendieck duality gives that then $$R\pi_*\omega_Y^\cdot\simeq_{qis}\omega_X^\cdot$$ Now if $X$ is CM, then the right hand side has only one non-zero cohomology. Now one could write $\omega_Y^\cdot$ as $$\omega_Y[n]\to \omega_Y^\cdot \to \omega_Y^+ \to^{+1} $$ So, if we also knew that $R^i\pi_*\omega_Y=0$ for $i>0$, then it would follow that $R\pi_*\omega_Y^+=0$ and I think that should imply that actually $\omega_Y^+=0$ which would imply that $Y$ is CM. So, this seems like a condition:

If $R^i\pi_*\mathcal O_Y=0$ and $R^i\pi_*\omega_Y=0$ for $i>0$, then what you want might follow. Then again, this might be more then what you would want to assume.

Any thoughts?