Lemma 45.4.1. The category $M_ k$ whose objects are motives over $k$ and morphisms are morphisms of motives over $k$ is a $\mathbf{Q}$-linear category. There is a contravariant functor

defined by $h(X) = (X, 1, 0)$ and $h(f) = [\Gamma _ f]$.

We fix a base field $k$. In this section we construct an additive Karoubian $\mathbf{Q}$-linear category $M_ k$ endowed with a symmetric monoidal structure and a contravariant functor

\[ h : \{ \text{smooth projective schemes over }k\} \longrightarrow M_ k \]

which maps products to tensor products and disjoint unions to direct sums. Our construction will be characterized by the fact that $h$ factors through the symmetric monoidal category whose objects are smooth projective varieties and whose morphisms are correspondences of degree $0$ such that the image of the projector $c_2$ on $h(\mathbf{P}^1_ k)$ from Example 45.3.7 is invertible in $M_ k$, see Lemma 45.4.8. At the end of the section we will show that every motive, i.e., every object of $M_ k$ to has a (left) dual, see Lemma 45.4.10.

A *motive* or a *Chow motive* over $k$ will be a triple $(X, p, m)$ where

$X$ is a smooth projective scheme over $k$,

$p \in \text{Corr}^0(X, X)$ satisfies $p \circ p = p$,

$m \in \mathbf{Z}$.

Given a second motive $(Y, q, n)$ we define a *morphism of motives* or a *morphism of Chow motives* to be an element of

\[ \mathop{\mathrm{Hom}}\nolimits ((X, p, m), (Y, q, n)) = q \circ \text{Corr}^{n - m}(X, Y) \circ p \subset \text{Corr}^{n - m}(X, Y) \]

Composition of morphisms of motives is defined using the composition of correspondences defined above.

Lemma 45.4.1. The category $M_ k$ whose objects are motives over $k$ and morphisms are morphisms of motives over $k$ is a $\mathbf{Q}$-linear category. There is a contravariant functor

\[ h : \{ \text{smooth projective schemes over }k\} \longrightarrow M_ k \]

defined by $h(X) = (X, 1, 0)$ and $h(f) = [\Gamma _ f]$.

**Proof.**
Follows immediately from Lemma 45.3.4.
$\square$

Lemma 45.4.2. The category $M_ k$ is Karoubian.

**Proof.**
Let $M = (X, p, m)$ be a motive and let $a \in \mathop{\mathrm{Mor}}\nolimits (M, M)$ be a projector. Then $a = a \circ a$ both in $\mathop{\mathrm{Mor}}\nolimits (M, M)$ as well as in $\text{Corr}^0(X, X)$. Set $N = (X, a, m)$. Since we have $a = p \circ a \circ a$ in $\text{Corr}^0(X, X)$ we see that $a : N \to M$ is a morphism of $M_ k$. Next, suppose that $b : (Y, q, n) \to M$ is a morphism such that $(1 - a) \circ b = 0$. Then $b = a \circ b$ as well as $b = b \circ q$. Hence $b$ is a morphism $b : (Y, q, n) \to N$. Thus we see that the projector $1 - a$ has a kernel, namely $N$ and we find that $M_ k$ is Karoubian, see Homology, Definition 12.4.1.
$\square$

We define a functor

\[ \otimes : M_ k \times M_ k \longrightarrow M_ k \]

On objects we use the formula

\[ (X, p, m) \otimes (Y, q, n) = (X \times Y, p \otimes q, m + n) \]

On morphisms, we use

\[ \xymatrix{ \mathop{\mathrm{Mor}}\nolimits ((X, p, m), (Y, q, n)) \times \mathop{\mathrm{Mor}}\nolimits ((X', p', m'), (Y', q', n')) \ar[d] \\ \mathop{\mathrm{Mor}}\nolimits ( (X \times X', p \otimes p', m + m'), (Y \times Y', q \otimes q', n + n')) } \]

given by the rule $(a, a') \longmapsto a \otimes a'$ where $\otimes $ on correspondences is as in Section 45.3. This makes sense: by definition of morphisms of motives we can write $a = q \circ c \circ p$ and $a' = q' \circ c' \circ p'$ with $c \in \text{Corr}^{n - m}(X, Y)$ and $c' \in \text{Corr}^{n' - m'}(X', Y')$ and then we obtain

\[ a \otimes a' = (q \circ c \circ p) \otimes (q' \circ c' \circ p') = (q \otimes q') \circ (c \otimes c') \circ (p \otimes p') \]

which is indeed a morphism of motives from $(X \times X', p \otimes p', m + m')$ to $(Y \times Y', q \otimes q', n + n')$.

Lemma 45.4.3. The category $M_ k$ with tensor product defined as above is symmetric monoidal with the obvious associativity and commutativity constraints and with unit $\mathbf{1} = (\mathop{\mathrm{Spec}}(k), 1, 0)$.

**Proof.**
Follows readily from Lemma 45.3.8. Details omitted.
$\square$

The motives $\mathbf{1}(n) = (\mathop{\mathrm{Spec}}(k), 1, n)$ are useful. Observe that

\[ \mathbf{1} = \mathbf{1}(0) \quad \text{and}\quad \mathbf{1}(n + m) = \mathbf{1}(n) \otimes \mathbf{1}(m) \]

Thus tensoring with $\mathbf{1}(1)$ is an autoequivalence of the category of motives. Given a motive $M$ we sometimes write $M(n) = M \otimes \mathbf{1}(n)$. Observe that if $M = (X, p, m)$, then $M(n) = (X, p, m + n)$.

Lemma 45.4.4. With notation as in Example 45.3.7

the motive $(X, c_0, 0)$ is isomorphic to the motive $\mathbf{1} = (\mathop{\mathrm{Spec}}(k), 1, 0)$.

the motive $(X, c_2, 0)$ is isomorphic to the motive $\mathbf{1}(-1) = (\mathop{\mathrm{Spec}}(k), 1, -1)$.

**Proof.**
We will use Lemma 45.3.4 without further mention. The structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ gives a correspondence $a \in \text{Corr}^0(\mathop{\mathrm{Spec}}(k), X)$. On the other hand, the rational point $x$ is a morphism $\mathop{\mathrm{Spec}}(k) \to X$ which gives a correspondence $b \in \text{Corr}^0(X, \mathop{\mathrm{Spec}}(k))$. We have $b \circ a = 1$ as a correspondence on $\mathop{\mathrm{Spec}}(k)$. The composition $a \circ b$ corresponds to the graph of the composition $X \to x \to X$ which is $c_0 = [x \times X]$. Thus $a = a \circ b \circ a = c_0 \circ a$ and $b = a \circ b \circ a = b \circ c_0$. Hence, unwinding the definitions, we see that $a$ and $b$ are mutually inverse morphisms $a : (\mathop{\mathrm{Spec}}(k), 1, 0) \to (X, c_0, 0)$ and $b : (X, c_0, 0) \to (\mathop{\mathrm{Spec}}(k), 1, 0)$.

We will proceed exactly as above to prove the second statement. Denote

\[ a' \in \text{Corr}^1(\mathop{\mathrm{Spec}}(k), X) = \mathop{\mathrm{CH}}\nolimits ^1(X) \]

the class of the point $x$. Denote

\[ b' \in \text{Corr}^{-1}(X, \mathop{\mathrm{Spec}}(k)) = \mathop{\mathrm{CH}}\nolimits _1(X) \]

the class of $[X]$. We have $b' \circ a' = 1$ as a correspondence on $\mathop{\mathrm{Spec}}(k)$ because $[x] \cdot [X] = [x]$ on $X = \mathop{\mathrm{Spec}}(k) \times X \times \mathop{\mathrm{Spec}}(k)$. Computing the intersection product $\text{pr}_{12}^*b' \cdot \text{pr}_{23}^*a'$ on $X \times \mathop{\mathrm{Spec}}(k) \times X$ gives the cycle $X \times \mathop{\mathrm{Spec}}(k) \times x$. Hence the composition $a' \circ b'$ is equal to $c_2$ as a correspondence on $X$. Thus $a' = a' \circ b \circ a' = c_2 \circ a'$ and $b' = b' \circ a' \circ b' = b' \circ c_2$. Recall that

\[ \mathop{\mathrm{Mor}}\nolimits ((\mathop{\mathrm{Spec}}(k), 1, -1), (X, c_2, 0)) = c_2 \circ \text{Corr}^1(\mathop{\mathrm{Spec}}(k), X) \subset \text{Corr}^1(\mathop{\mathrm{Spec}}(k), X) \]

and

\[ \mathop{\mathrm{Mor}}\nolimits ((X, c_2, 0), (\mathop{\mathrm{Spec}}(k), 1, -1)) = \text{Corr}^{-1}(X, \mathop{\mathrm{Spec}}(k)) \circ c_2 \subset \text{Corr}^{-1}(X, \mathop{\mathrm{Spec}}(k)) \]

Hence, we see that $a'$ and $b'$ are mutually inverse morphisms $a' : (\mathop{\mathrm{Spec}}(k), 1, -1) \to (X, c_0, 0)$ and $b' : (X, c_0, 0) \to (\mathop{\mathrm{Spec}}(k), 1, -1)$. $\square$

Remark 45.4.5 (Lefschetz and Tate motive). Let $X = \mathbf{P}^1_ k$ and $c_2$ be as in Example 45.3.7. In the literature the motive $(X, c_2, 0)$ is sometimes called the *Lefschetz motive* and depending on the reference the notation $L$, $\mathbf{L}$, $\mathbf{Q}(-1)$, or $h^2(\mathbf{P}^1_ k)$ may be used to denote it. By Lemma 45.4.4 the Lefschetz motive is isomorphic to $\mathbf{1}(-1)$. Hence the Lefschetz motive is invertible (Categories, Definition 4.43.4) with inverse $\mathbf{1}(1)$. The motive $\mathbf{1}(1)$ is sometimes called the *Tate motive* and depending on the reference the notation $L^{-1}$, $\mathbf{L}^{-1}$, $\mathbf{T}$, or $\mathbf{Q}(1)$ may be used to denote it.

Lemma 45.4.6. The category $M_ k$ is additive.

**Proof.**
Let $(Y, p, m)$ and $(Z, q, n)$ be motives. If $n = m$, then a direct sum is given by $(Y \amalg Z, p + q, m)$, with obvious notation. Details omitted.

Suppose that $n < m$. Let $X$, $c_2$ be as in Example 45.3.7. Then we consider

\begin{align*} (Z, q, n) & = (Z, q, m) \otimes (\mathop{\mathrm{Spec}}(k), 1, -1) \otimes \ldots \otimes (\mathop{\mathrm{Spec}}(k), 1, -1) \\ & \cong (Z, q, m) \otimes (X, c_2, 0) \otimes \ldots \otimes (X, c_2, 0) \\ & \cong (Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m) \end{align*}

where we have used Lemma 45.4.4. This reduces us to the case discussed in the first paragraph. $\square$

Lemma 45.4.7. In $M_ k$ we have $h(\mathbf{P}^1_ k) \cong \mathbf{1} \oplus \mathbf{1}(-1)$.

**Proof.**
This follows from Example 45.3.7 and Lemma 45.4.4.
$\square$

Lemma 45.4.8. Let $X$, $c_2$ be as in Example 45.3.7. Let $\mathcal{C}$ be a $\mathbf{Q}$-linear Karoubian symmetric monoidal category. Any $\mathbf{Q}$-linear functor

\[ F : \left\{ \begin{matrix} \text{smooth projective schemes over }k
\\ \text{morphisms are correspondences of degree }0
\end{matrix} \right\} \longrightarrow \mathcal{C} \]

of symmetric monoidal categories such that the image of $F(c_2)$ on $F(X)$ is an invertible object, factors uniquely through a functor $F : M_ k \to \mathcal{C}$ of symmetric monoidal categories.

**Proof.**
Denote $U$ in $\mathcal{C}$ the invertible object which is assumed to exist in the statement of the lemma. We extend $F$ to motives by setting

\[ F(X, p, m) = \left(\text{the image of the projector }F(p)\text{ in }F(X)\right) \otimes U^{\otimes -m} \]

which makes sense because $U$ is invertible and because $\mathcal{C}$ is Karoubian. An important feature of this choice is that $F(X, c_2, 0) = U$. Observe that

\begin{align*} F((X, p, m) \otimes (Y, q, n)) & = F(X \times Y, p \otimes q, m + n) \\ & = \left(\text{the image of }F(p \otimes q)\text{ in }F(X \times Y)\right) \otimes U^{\otimes -m - n} \\ & = F(X, p, m) \otimes F(Y, q, n) \end{align*}

Thus we see that our rule is compatible with tensor products on the level of objects (details omitted).

Next, we extend $F$ to morphisms of motives. Suppose that

\[ a \in \mathop{\mathrm{Hom}}\nolimits ((Y, p, m), (Z, q, n)) = q \circ \text{Corr}^{n - m}(Y, Z) \circ p \subset \text{Corr}^{n - m}(Y, Z) \]

is a morphism. If $n = m$, then $a$ is a correspondence of degree $0$ and we can use $F(a) : F(Y) \to F(Z)$ to get the desired map $F(Y, p, m) \to F(Z, q, n)$. If $n < m$ we get canonical identifications

\begin{align*} s : F((Z, q, n)) & \to F(Z, q, m) \otimes U^{m - n} \\ & \to F(Z, q, m) \otimes F(X, c_2, 0) \otimes \ldots \otimes F(X, c_2, 0) \\ & \to F((Z, q, m) \otimes (X, c_2, 0) \otimes \ldots \otimes (X, c_2, 0)) \\ & \to F((Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m)) \end{align*}

Namely, for the first isomorphism we use the definition of $F$ on motives above. For the second, we use the choice of $U$. For the third we use the compatibility of $F$ on tensor products of motives. The fourth is the definition of tensor products on motives. On the other hand, since we similarly have an isomorphism

\[ \sigma : (Z, q, n) \to (Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m) \]

(see proof of Lemma 45.4.6). Composing $a$ with this isomorphism gives

\[ \sigma \circ a \in \mathop{\mathrm{Hom}}\nolimits ((Y, p, m), (Z \times X^{m - n}, q \otimes c_2 \otimes \ldots \otimes c_2, m)) \]

Putting everything together we obtain

\[ s^{-1} \circ F(\sigma \circ a) : F(Y, p, m) \to F(Z, q, n) \]

If $n > m$ we similarly define isomorphisms

\[ t : F((Y, p, m)) \to F((Y \times X^{n - m}, p \otimes c_2 \otimes \ldots \otimes c_2, n)) \]

and

\[ \tau : (Y, p, m)) \to (Y \times X^{n - m}, p \otimes c_2 \otimes \ldots \otimes c_2, n) \]

and we set $F(a) = F(a \circ \tau ^{-1}) \circ t$. We omit the verification that this construction defines a functor of symmetric monoidal categories. $\square$

Lemma 45.4.9. Let $X$ be a smooth projective scheme over $k$ which is equidimensional of dimension $d$. Then $h(X)(d)$ is a left dual to $h(X)$ in $M_ k$.

**Proof.**
We will use Lemma 45.3.1 without further mention. We compute

\[ \mathop{\mathrm{Hom}}\nolimits (\mathbf{1}, h(X) \otimes h(X)(d)) = \text{Corr}^ d(\mathop{\mathrm{Spec}}(k), X \times X) = \mathop{\mathrm{CH}}\nolimits ^ d(X \times X) \]

Here we have $\eta = [\Delta ]$. On the other hand, we have

\[ \mathop{\mathrm{Hom}}\nolimits (h(X)(d) \otimes h(X), \mathbf{1}) = \text{Corr}^{-d}(X \times X, \mathop{\mathrm{Spec}}(k)) = \mathop{\mathrm{CH}}\nolimits _ d(X \times X) \]

and here we have the class $\epsilon = [\Delta ]$ of the diagonal as well. The composition of the correspondence $[\Delta ] \otimes 1$ with $1 \otimes [\Delta ]$ either way is the correspondence $[\Delta ] = 1$ in $\text{Corr}^0(X, X)$ which proves the required diagrams of Categories, Definition 4.43.5 commute. Namely, observe that

\[ [\Delta ] \otimes 1 \in \text{Corr}^ d(X, X \times X \times X) = \mathop{\mathrm{CH}}\nolimits ^{2d}(X \times X \times X \times X) \]

is given by the class of the cycle $\text{pr}^{1234, -1}_{23}(\Delta ) \cap \text{pr}^{1234, -1}_{14}(\Delta )$ with obvious notation. Similarly, the class

\[ 1 \otimes [\Delta ] \in \text{Corr}^{-d}(X \times X \times X, X) = \mathop{\mathrm{CH}}\nolimits ^{2d}(X \times X \times X \times X) \]

is given by the class of the cycle $\text{pr}^{1234, -1}_{23}(\Delta ) \cap \text{pr}^{1234, -1}_{14}(\Delta )$. The composition $(1 \otimes [\Delta ]) \circ ([\Delta ] \otimes 1)$ is by definition the pushforward $\text{pr}^{12345}_{15, *}$ of the intersection product

\[ [\text{pr}^{12345, -1}_{23}(\Delta ) \cap \text{pr}^{12345, -1}_{14}(\Delta )] \cdot [\text{pr}^{12345, -1}_{34}(\Delta ) \cap \text{pr}^{12345, -1}_{15}(\Delta )] = [\text{small diagonal in } X^5] \]

which is equal to $\Delta $ as desired. We omit the proof of the formula for the composition in the other order. $\square$

Lemma 45.4.10. Every object of $M_ k$ has a left dual.

**Proof.**
Let $M = (X, p, m)$ be an object of $M_ k$. Then $M$ is a summand of $(X, 0, m) = h(X)(m)$. By Homology, Lemma 12.17.3 it suffices to show that $h(X)(m) = h(X) \otimes \mathbf{1}(m)$ has a dual. By construction $\mathbf{1}(-m)$ is a left dual of $\mathbf{1}(m)$. Hence it suffices to show that $h(X)$ has a left dual, see Categories, Lemma 4.43.8. Let $X = \coprod X_ i$ be the decomposition of $X$ into irreducible components. Then $h(X) = \bigoplus h(X_ i)$ and it suffices to show that $h(X_ i)$ has a left dual, see Homology, Lemma 12.17.2. This follows from Lemma 45.4.9.
$\square$

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