Showing posts with label projective. Show all posts
Showing posts with label projective. Show all posts

Wednesday, March 15, 2017

Lengths of paths on projective varieties

This post contains calculations that continue on the ideas from the previous post "Fubini--Study metric," 2017-03-05. First we suppose that $\gamma$ lies on a curve $C\subset \P^2$, with the curve defined as the zero locus of a polynomial $P$. Taking the derivative of $P$ on $\C^2$ gives $P_{z_1}dz_1 + P_{z_2}dz_2=0$, which can be manipulated to give
\begin{align*}
dz_2 & = \frac{-P_{z_1}}{P_{z_2}}dz_1, & \frac\dy{\dy z_2} & = \frac{-P_{z_2}}{P_{z_1}} \frac\dy{\dy z_1},\\
d\overline{z_2} & = \frac{-\overline{P_{z_1}}}{\overline{P_{z_2}}}d\overline{z_1}, & \frac\dy{\dy \overline{z_2}} & = \frac{-\overline{P_{z_2}}}{\overline{P_{z_1}}} \frac\dy{\dy \overline{z_1}}.
\end{align*}
Using the above and an equation from the mentioned post, for $e = \frac\dy{\dy z_1} + \frac\dy{\dy \overline {z_1}} + \frac\dy{\dy z_2} + \frac\dy{\dy \overline{z_2}}$, we get
\begin{align*}
\frac{d \gamma}{dt} & = \left(\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2'\right)\frac\dy{\dy z_1} + \left(\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right)\frac\dy{\dy \overline{z_1}} \\
\left(\sum_{k,\ell=1}^2\chi_{k\ell}(\gamma)dz_k\wedge d\overline{z_\ell}\right)(e,e) & = 1+|\gamma_2|^2 + \frac{\overline{P_{z_1}}}{\overline{P_{z_2}}} \overline \gamma_1\gamma_2 + \frac{P_{z_1}}{P_{z_2}}\gamma_1\overline\gamma_2 + \left|\frac{P_{z_1}}{P_{z_2}}\right|^2 \left(1+|\gamma_1|^2\right) = 1 +\left|\frac{P_{z_1}}{P_{z_2}}\right|^2 + \left|\frac{P_{z_1}}{P_{z_2}}\gamma_1+\gamma_2\right|^2, \\
(dz_1\wedge d\overline{z_1})\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right) & = \det
\begin{bmatrix}
\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2' & i\left(\overline\gamma_1'-\frac{P_{z_2}}{P_{z_1}}\overline \gamma_2'\right) \\[5pt]
\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2' & -i\left(\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right)
\end{bmatrix} = -2i \left|\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right|^2.
\end{align*}
Hence
\[
g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right) = \frac{\left(1 +\left|\frac{P_{z_1}}{P_{z_2}}\right|^2 + \left|\frac{P_{z_1}}{P_{z_2}}\gamma_1+\gamma_2\right|^2\right)\left|\gamma_1' - \frac{\overline{P_{z_2}}}{\overline{P_{z_1}}}\gamma_2'\right|^2}{\pi\left(1+|\gamma_1|^2+|\gamma_2|^2\right)^2}.
\]

Now we move to $\P^n$, and consider $X\subset \P^n$ a complete intersection of codimension $r$, or the zero set of polynomials $P_1=0,\dots,P_r=0$. Expressing some covectors in terms of others reduces the number of determinants we calculated above from $2n$ to $2(n-r)$. Then
\begin{align*}
P_{1,z_1}dz_1 + \cdots + P_{1,z_n}dz_n & = 0, & dz_n & = c_{n,1}dz_1 + \cdots + c_{n,n-r}dz_{n-r}, \\
& \ \ \vdots & & \ \ \vdots \\
P_{r,z_1}dz_1 + \cdots + P_{r,z_n}dz_n & = 0, & dz_{n-r+1} & = c_{n-r+1,1}dz_1 + \cdots + c_{n-r+1,n-r}dz_{n-r},
\end{align*}
for the $c_{i,j}$ some combinations of the $P_{k,z_\ell}$. By orthonormality of the basis vectors, and assuming that the $c_{i,j}$ are all non-zero, we find
\[
\frac\dy{\dy z_i} = \sum_{j=1}^{n-r} \frac1{(n-r)c_{i,j}}\frac\dy{\dy z_j},\hspace{2cm}
\frac\dy{\dy \overline{z_i}} = \sum_{j=1}^{n-r} \frac1{(n-r)\overline{c_{i,j}}}\frac\dy{\dy \overline{z_j}},
\]
for all integers $n-r<i\leqslant n$. This allows us to rewrite the path derivative as
\begin{align*}
\frac{d\gamma}{dt} & = \sum_{i=1}^n \overline \gamma_i'\frac\dy{\dy z_i} +\gamma_i'\frac\dy{\dy \overline{z_i}} \\
& = \sum_{i=1}^{n-r} \left(\overline \gamma_i'\frac\dy{\dy z_i} +\gamma_i'\frac\dy{\dy \overline{z_i}}\right) +\sum_{i=n-r+1}^n \left(\sum_{j=1}^{n-r} \frac{\overline \gamma_i'}{(n-r)c_{i,j}}\frac\dy{\dy z_j} + \sum_{j=1}^{n-r} \frac{\gamma_i'}{(n-r)\overline{c_{i,j}}}\frac\dy{\dy \overline{z_j}}\right) \\
& = \sum_{i=1}^{n-r}\left(\overline\gamma_i' + \sum_{j=n-r+1}^n \frac{\overline\gamma_j'}{(n-r)c_{j,i}}\right)\frac\dy{\dy z_i} + \left(\gamma_i'+\sum_{j=n-r+1}^n \frac{\gamma_j'}{(n-r)\overline{c_{j,i}}}\right)\frac\dy{\dy \overline{z_i}}.
\end{align*}

In the case of a curve in $\P^n$, when $r=n-1$, let $c_{1,1}=1$ and  $e = \frac\dy{\dy z_1} + \frac\dy{\dy \overline {z_1}} + \cdots + \frac\dy{\dy z_n} + \frac\dy{\dy \overline{z_n}}$ to get
\begin{align*}
 \frac{d\gamma}{dt} & = \left(\sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}}\right)\frac\dy{\dy z_1} + \left(\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right)\frac\dy{\dy \overline{z_1}},\\
 \left(\sum_{k,\ell=1}^n\chi_{k\ell}(\gamma)dz_k\wedge d\overline{z_\ell}\right)(e,e) & = \sum_{k,\ell=1}^n \left(1+\sum_{i=1}^n |\gamma_i|^2\right)\delta_{k\ell} - \overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}, \\
(dz_1\wedge d\overline{z_1})\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right) & = \det
\begin{bmatrix}
\sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}} & i \sum_{j=1}^n \frac{\overline\gamma_j'}{c_{j1}} \\[5pt]
\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}} & -i\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}
\end{bmatrix} = -2i \left|\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right|^2.
\end{align*}
Hence
\[
g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right) = \frac{\left(\sum_{k,\ell=1}^n \left(1+\sum_{i=1}^n |\gamma_i|^2\right)\delta_{k\ell} - \overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}\right)\left|\sum_{j=1}^n \frac{\gamma_j'}{\overline{c_{j1}}}\right|^2}{\pi \left(1+\sum_{i=1}^n |\gamma_i|^2\right)^2}.
\]
The terms $\overline{\gamma_kc_{\ell1}}\gamma_\ell c_{k1}$ may be rearranged into terms $|\gamma_kc_{\ell1}-\gamma_\ell c_{k1}|^2$, but it does not provide any enlightening results, similarly to the rest of this post.

Sunday, March 5, 2017

The Fubini-Study metric and length in projective space

In this post we inspect how the Fubini-Study metric works and compute an example. Professor Mihai Paun for helpful discussions. Recall that from projective space $\P^n$ there are natural maps
\[
[x_0:x_1:\cdots:x_n]\tov{\vp_i}\left(\frac{x_0}{x_i},\dots,\widehat{\frac{x_i}{x_i}},\dots,\frac{x_n}{x_i}\right)
\]
for $i=0,\dots,n$. The maps land in $\C^n$ with coordinates $(z_1,z_2,\dots,z_n)$. We use $\vp_0$ as the main map, and conflate notation for objects in $\P^n$ and in $\C^n$ under $\vp_0$. Most of this post deals with the $n=2$ case.


The metric


The metric used on $\P^n$ is the Fubini-Study metric. Directly from Section 3.1 of Huybrechts, for $n=2$ the associated differential 2-form and its image in $\C^2$ are
\begin{align*}
\omega & = \frac i{2\pi}\partial \bar\partial \log\left(1+\left|\frac{x_1}{x_0}\right|^2+\left|\frac{x_2}{x_0}\right|^2\right), \\
\vp_0(\omega) & = \frac i{2\pi}\partial \bar\partial \log\left(1+\left|z_1\right|^2+\left|z_2\right|^2\right) \\
& =  \underbrace{\frac{i}{2\pi (1+|z_1|^2+|z_2|^2)^2}}_{\lambda_2}\sum_{k,\ell=1}^2\underbrace{(1+|z_1|^2+|z_2|^2)\delta_{k\ell} -\overline{z_k}z_\ell}_{\chi_{k\ell}}dz_k\wedge d\overline{z_\ell}. \hspace{1cm} (1)
\end{align*}
A Hermitian metric on a complex manifold $X$ may be described as a 2-tensor $h=g-i\omega$, where $g$ is a Riemannian metric (also a 2-tensor) on the underlying real manifold and $\omega$ is a Kahler form, a 2-form. As in Lemma 3.3 of Voisin, the relationship between $g$ and $\omega$ is given by
\[
g(u,v)=\omega(u,Iv)=\omega(Iu,v), \hspace{1cm} (2)
\]
where $I:T_xX\to T_xX$ is a tangent space endomorphism defined by
\[
\begin{array}{r c l}
I|_{T^{1,0}_xX} & = & i\cdot \id, \\
\frac{\dy}{\dy z_i} & \mapsto & i\frac{\dy}{\dy z_i},
\end{array}
\hspace{1cm}
\begin{array}{r c l}
I|_{T^{0,1}_xX} & = & -i\cdot \id, \\
\frac{\dy}{\dy \overline{z_i}} & \mapsto & -i\frac{\dy}{\dy \overline{z_i}},
\end{array}
\]
as in Proposition 1.3.1 of Huybrechts.

An application


Let $\gamma:[0,1]\to \C^2$ be a path, described as $\gamma(t)=(\gamma_1(t),\gamma_2(t))$. The derivative of $\gamma$ with respect to $t$, in the basis $\frac{\dy}{\dy z_1}$, $\frac{\dy}{\dy \overline{z_1}}$, $\frac{\dy}{\dy z_2}$, $\frac{\dy}{\dy \overline{z_2}}$ is given by
\[
\frac{d\gamma_1}{dt} = \frac{du_1}{dt}\frac\dy{\dy x_1} + i\frac{dv_1}{dt}\frac\dy{\dy y_1} = \frac{du_1}{dt}\left(\frac\dy{\dy \overline{z_1}}+\frac\dy{\dy z_1}\right) + i\frac{dv_1}{dt}\left(\frac\dy{\dy \overline{z_1}} -\frac{\dy}{\dy z_1}\right) = \underbrace{\left(\frac{du_1}{dt} + i\frac{dv_1}{dt}\right)}_{\gamma_1'}\frac\dy{\dy \overline {z_1}} + \underbrace{\left(\frac{du_1}{dt}-i\frac{dv_1}{dt}\right)}_{\overline \gamma_1'}\frac\dy{\dy z_1},\]
and analogously for $\gamma_2$. Hence
\[
\frac{d\gamma}{dt} =
\overline \gamma_1'\frac\dy{\dy z_1} + \gamma_1'\frac\dy{\dy \overline{z_1}} + \overline \gamma_2' \frac\dy{\dy z_2} + \gamma_2' \frac{\dy}{\dy \overline{z_2}}. \hspace{1cm} (3)
\]
The length of $\gamma$ is
\[
\int_0^1\sqrt{g\left(\frac{d\gamma}{dt},\frac{d\gamma}{dt}\right)}\ dt = \int_0^1\sqrt{\omega\left(\frac{d\gamma}{dt},I\frac{d\gamma}{dt}\right)}\ dt,
\]
using equation (2). Recall that the pairing of vectors with covectors is given by\[
\left(d\alpha_1\wedge \cdots \wedge d\alpha_n\right)\left(\frac\dy{\dy \beta_1},\dots,\frac\dy{\dy \beta_n}\right) = \det\begin{bmatrix}
d\alpha_1\frac\dy{\dy \beta_1} & d\alpha_1\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_1\frac\dy{\dy \beta_n} \\
d\alpha_2\frac\dy{\dy \beta_1} & d\alpha_2\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_2\frac\dy{\dy \beta_n} \\
\vdots & \vdots & \ddots & \vdots \\
d\alpha_n\frac\dy{\dy \beta_1} & d\alpha_n\frac{\dy}{\dy \beta_2} & \cdots & d\alpha_n\frac\dy{\dy \beta_n}
\end{bmatrix}
 \ \ = \ \
\det\left(d\alpha_i\frac\dy{\dy \beta_j}\right),
\]
for $\alpha_i,\beta_j$ a basis of the underlying real manifold (as in the previous post "Vector fields," 2016-10-10). The components of the vector (3) may be viewed as given in directions $z_1,\overline{z_1}, z_2,\overline{z_2}$, respectively, which also indicates how the coefficient functions $\chi_{k\ell}$ act on (3). Apply the definition of $\omega$ from equation (1), and note that we are always at the tangent space to the point $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, to get that
\begin{align*}
& \omega\left(\frac{d \gamma}{dt},I\frac{d\gamma}{dt}\right) \\
& = \lambda_2(\gamma(t)) \sum_{k,\ell=1}^2 \chi_{k\ell}(\gamma(t)) dz_k\wedge d\overline{z_\ell}\left(\overline \gamma_1'\frac\dy{\dy z_1} + \gamma_1'\frac\dy{\dy \overline{z_1}} + \overline \gamma_2' \frac\dy{\dy z_2} + \gamma_2' \frac{\dy}{\dy \overline{z_2}}, i\overline \gamma_1'\frac\dy{\dy z_1} - i\gamma_1'\frac\dy{\dy \overline{z_1}} + i\overline \gamma_2' \frac\dy{\dy z_2} -i\gamma_2' \frac{\dy}{\dy \overline{z_2}}\right) \\
& = \lambda_2(\gamma(t)) \sum_{k,\ell=1}^2 \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix} \\
& = \frac{(1+|\gamma_2(t)|^2)|\gamma_1'(t)|^2 - \overline\gamma_1(t)\gamma_2(t)\overline\gamma_1'(t)\gamma_2'(t) - \overline\gamma_2(t)\gamma_1(t)\overline \gamma_2'(t)\gamma_1'(t) + (1+|\gamma_1(t)|^2) |\gamma_2'(t)|^2}{\pi\left(1+\left|\gamma_1(t)\right|^2+\left|\gamma_2(t)\right|^2\right)^2}.\end{align*}
Unfortunately this expression does not simplify too much. In $\P^n$, with $\gamma = (\gamma_1,\dots,\gamma_n):[0,1]\to \C^n$, we have that
\[
g\left(\frac{d \gamma}{dt},\frac{d\gamma}{dt}\right) = \lambda_n(\gamma(t)) \sum_{k,\ell=1}^n \chi_{k\ell}(\gamma(t))\det
\begin{bmatrix}
\overline \gamma_k'(t) & i\overline \gamma_k'(t) \\[5pt] \gamma_\ell'(t) & -i\gamma_\ell'(t)
\end{bmatrix}.
\]

An example


Here we compute the distance between two points in $\P^2$. Let $\gamma$ be the straight line segment connecting $p=[p_0:p_1:p_2]$ and $q=[q_0:q_1:q_2]$. The word "straight" is used loosely, and means the segment may be parametrized as
\[
\gamma(t) = [(1-t)p_0+tq_0:(1-t)p_1+tq_1:(1-t)p_2+tq_2],
\]
so $\gamma(0)=p$ and $\gamma(1)=q$. The image of $\gamma$ under $\vp_0$ and its derivative are given by
\[
\vp_0(\gamma(t)) = \left(\frac{(1-t)p_1+tq_1}{(1-t)p_0+tq_0}, \frac{(1-t)p_2+tq_2}{(1-t)p_0+tq_0}\right) = (\gamma_1,\gamma_2),
\hspace{2cm}
\gamma_i' = \frac{q_ip_0-q_0p_i}{((1-t)p_0+tq_0)^2}.
\]
If, for example, $p=[1:1:0]$ and $q=[1:0:1]$, then
\[
\text{length}(\gamma) = \frac{3}{4\pi}\int_0^1\frac1{(t^2-t+1)^2}\ dt = \frac{9+2\pi\sqrt 3}{18\pi}.
\]

A further goal is to consider the path $\gamma$ as lying on a projective variety, beginning with a complete intersection. This would allow some of the $dz_i$ to be expressed in terms of other $dz_j$.

References: Huybrechts (Complex geometry, Section 3.1), Voisin (Hodge theory and complex algebraic geometry 1, Chapter 3.1), Wells (Differential analysis on complex manifolds, Chapter V.4)

Tuesday, June 28, 2016

The conditioning number of a projective curve

Let $C$ be a smooth algebraic curve in $\P^2$. That is, for some homogeneous $f\in \C[x_0,x_1,x_2]$ we let $C = \{x\in \P^2\ :\ f(x)=0\}$. Describe $C$ as a manifold via the usual open sets $U_i = \{x\in \P^2\ :\ x_i\neq 0\}$ and charts
\[
\begin{array}{r c l}
\varphi_0\ :\ U_0 & \to & \C^2, \\\
[x_0:x_1:x_2] & \mapsto & (\frac{x_1}{x_0},\frac{x_2}{x_0}),
\end{array}
\hspace{1cm}
\begin{array}{r c l}
\varphi_1\ :\ U_1 & \to & \C^2, \\\
[x_0:x_1:x_2] & \mapsto & (\frac{x_0}{x_1},\frac{x_2}{x_1}),
\end{array}
\hspace{1cm}
\begin{array}{r c l}
\varphi_2\ :\ U_2 & \to & \C^2, \\\
[x_0:x_1:x_2] & \mapsto & (\frac{x_0}{x_2},\frac{x_1}{x_2}).
\end{array}
\]
Let $w=[w_0:w_1:w_2]\in \P^2$ for which $f(w)=0$. The Jacobian of $C$ at $w$ is then
\[
J_w = \left[
\left.\frac{\dy f}{\dy x_0}\right|_w \ :\  \left.\frac{\dy f}{\dy x_1}\right|_w \ :\  \left.\frac{\dy f}{\dy x_2}\right|_w
\right] \in \P^2.
\]
Assume that $\left.\frac{\dy f}{\dy x_0}\right|_w\neq 0$ and pass to $\varphi_0(U_0)$ to get the Jacobian to be
\[
J_w^0 = \left(
\frac{\dy f/\dy x_1|_w}{\dy f/\dy x_0|_w}\ ,\ \frac{\dy f/\dy x_2|_w}{\dy f/\dy x_0|_w}\right)  \in \C^2.
\]
Assume that $w_0\neq 0$, so the tangent line to $\varphi_0(C)\subset \C^2$ at $\varphi_0(w)=(w_1/w_0,w_2/w_0)$ is
\[
T_{\varphi_0(w)}= \{\varphi_0(w)+tJ_w^0\ :\ t\in \C\}\subset \C^2.
\]
A vector orthogonal to the Jacobian $J_w^0$ is
\[
\bar J_w^0 = \left(-\frac{\dy f/\dy x_2|_w}{\dy f/\dy x_0|_w}\ ,\ \frac{\dy f/\dy x_1|_w}{\dy f/\dy x_0|_w}\right) \in \C^2,
\]
so the space space normal to $T_{\varphi_0(w)}$ is given by
\[
T_{\varphi_0(w)}^\perp = \{\varphi_0(w)+t\bar J_w^0\ :\ t\in \C\}\subset \C^2.
\]

Example: Let $C\subset \P^2$ be the zero locus of $f(x_0,x_1,x_2) = x_0^2+x_1x_2-x_1x_0$. The Jacobian is $J = [2x_0-x_1:x_2-x_0:x_1]$, and as $J=0$ implies $x_0=x_1=x_2=0$, but $0\not\in\P^2$, the curve $C$ is smooth. Consider two points $w=[1:1:0],z=[2:1:-2]\in C$, at which the Jacobian is
\[
J_w = [1:-1:1]
\hspace{1cm},\hspace{1cm}
J_z = [3:-4:1].
\]
Both $w_0$ and $z_0$ are non-zero, with $\varphi_0(w)=(1,0)$ and $\varphi_0(z)=(1/2,-1)$, giving the tangent and normal spaces to be
\begin{align*}
T_{(1,0)} & = \{(1,0)+t(-1,1)\ :\ t\in \C\}, & T_{(1/2,-1)} & = \{(1/2,-1)+s(-4/3,1/3)\ :\ s\in \C\}, \\
T^\perp_{(1,0)} & = \{(1,0)+t(-1,-1)\ :\ t\in \C\}, & T_{(1/2,-1)}^\perp & = \{(1/2,-1)+s(-1/3,-4/3)\ :\ s\in \C\}.
\end{align*}
The two normal spaces intersect at $(t,s)=(1/3,-1/2)$ at distances of $1/3\cdot ||(-1,-1)|| = \sqrt 2/3\approx 0.471$ and $1/2\cdot||(-1/3,-4/3)|| = \sqrt{17}/3\approx 1.374$ from the points $\varphi_0(w),\varphi_0(z)$, respectively. Hence the conditioning number of $C$ is at most $\sqrt 2/3$.

Given a smooth projective curve and a finite set of points, this Sage code will calculate the conditioning number from that collection of points.

Sunday, March 20, 2016

Exactness and derived functors

 Lecture topic

Let $0\to X\to Y\to Z\to 0$ be a short exact sequence of objects in a category $A$. Let $\mathcal F:A\to B$ be a covariant functor.

Definition:
The functor $\mathcal F$ is right-exact if $\mathcal F(X)\to\mathcal F(Y)\to \mathcal F(Z)\to 0$ is an exact sequence. It is left-exact if $0\to \mathcal F(X)\to\mathcal F(Y)\to \mathcal F(Z)$ is an exact sequence. It is exact if it is both left- and right-exact.

Example: These are some examples of left- and right-exact functors:
    $\Hom_A(X,-)$ is covariant left-exact
    $\Hom_A(-,X)$ is contravariant left-exact
    $-\otimes_R X$ is covariant right-exact, for $X$ a left $R$-module

Recall that $X\otimes_R Y$ is naturally isomorphic to $Y\otimes_RX$.

Definition: An object $X\in \Obj(A)$ is projective if $\Hom_A(X,-)$ is an exact functor. Similarly, $X$ is injective if $\Hom_A(-,X)$ is an exact functor.

Recall that a projective resolution of an object $X$ is a sequence of projective objects $\cdots\to P_2\to P_1\to P_0$ that may or may not terminate on the left. The homology of the sequence in degree 0 is $X$, and trivial in other degrees. Similarly, an injective resolution of $X$ is a sequence of injective objects $I_0\to I_1\to I_2\to\cdots$ that may or may not terminate on the right. The cohomology is also concentrated in degree 0, and is $X$ there. A free resolution is a projective resolution where all the objects are free (whatever that means in the context).

These types of resolutions may not exist. A category "has enough injectives (projectives)" means we can always construct injective (projective) resolutions.

Definition: Let $\mathcal F:A\to B$ be a covariant right-exact functor and $\mathcal G:A\to B$ a covariant left-exact functor. Let $X\in \Obj(A)$ with $P_\bullet$ a projective resolution of $X$ and $I_\bullet$ an injective resolution of $X$. The $i$th left-derived functor of $\mathcal F$ is $L_i\mathcal F(X) = H_i(\mathcal F(P_\bullet))$. The $i$th right-derived functor of $\mathcal G$ is $R^i\mathcal G(X) = H^i(\mathcal G(I_\bullet))$.

These objects of $B$ are well-defined up to natural isomorphism. Note that $\mathcal F^{op}:A^{op}\to B^{op}$ is a contravariant right-exact functor. Moreover, if $\mathcal F$ was contravariant right-exact and $\mathcal G$ was contravariant left-exact, then $L_i\mathcal F(X)=H_i(\mathcal F(I_\bullet))$ and $R^i\mathcal G(X)=H^i(\mathcal G(P_\bullet))$.

Example:
Let $R$ be a ring with $X$ and $Y$ both $R$-bimodules. Then
\begin{align*}
\Tor_i^R(Y,X) & =  L_i(-\otimes_RX)(Y) &
\Ext^i_R(X,Y) & = R^i(\Hom_R(X,-))(Y) \\
& = L_i(Y\otimes_R - )(X),
&& = R^i(\Hom_R(-,Y))(X).
\end{align*}
Recall that $\Tor_i^R(Y,X)$ is canonically isomorphic to $\Tor_i^R(X,Y)$, but it is not true for $\Ext$. Also note that $\Hom_R(X,-)$ is covariant and $\Hom_R(-,Y)$ is contravariant, while $-\otimes_R X$ and $Y\otimes_R -$ are both covariant functors.
References: Weibel (An introduction to homological algebra, Chapter 2)