From what I can gather, an algebraic torus is an algebraic group defined over a field, which is isomorphic to ~something~. I can't quite tell based on the definitions below what that something is exactly.

  1. Wondering if one could write a formal definition of an algebraic torus that explains some of the components in a little more detail.
  2. Wondering what the $\times$ is in $\mathbf {T} ({\overline {F}})\cong ({\overline {F}}^{\times })^{r}$.

The Wikipedia one (4) has a few notation symbols I haven't seen before. (3) sort of makes sense but has a few notational symbols new as well. (2) provides some additional aspects. (1) is short but I also don't follow the notation. (5) is the closest to making sense.

(1) Let $k$ be a field. An algebraic $k$-torus $T$ is an algebraic $k$-group such that over a (fixed) separable closure $\bar{k}$ of $k$ it becomes isomorphic to a direct product of $d$ copies of the multiplicative group:

$$T \times_k \bar{k} \approx \mathbb{G}^d_{m,\bar{k}}$$

(here $d$ is the dimension of $T$).

(2) Fix a field $k$, and let $k_s$ be a fixed separable closure of $k$. Let $\mathbb{A}^d$ denote $d$-dimensional affine space and let $\mathbb{G}_m$ denote the multiplicative group. If $V$ is a variety and $D$ is a finite set, write

$$V^D := \bigoplus_{\delta\in D} V \approx V^{|D|}$$

If $D$ is a group, then $D$ acts on $V^D$ by permuting the summands. Write

$$\mathbb{A}^D := (A^1)^D = \bigoplus_{\delta\in D}\mathbb{A}^1$$

If $G$ is a group and $H$ is a subgroup, define a norm map $\mathbf{N}_H : \mathbb{G}^G_m \to \mathbb{G}^{G/H}_m$ by

$$(\alpha g)_{g\in G} \mapsto (\prod_{\gamma \in gH} \alpha_\gamma)_{gH\in G/H}$$

and let

$$\mathbb{T}_G := \ker[\mathbb{G}^G_m \overset{\oplus\mathbf{N}_H}{\longrightarrow} \bigoplus_{1\neq H \subseteq G} \mathbb{G}^{G/H}_m]$$

An algebraic torus $T$ (over $k$) is an algebraic group defined over $k$ that is isomorphic over $k_s$ to $\mathbb{G}^d_m$, where $d$ is necessarily the dimension of $T$. If $k \subseteq L \subseteq k_s$ and $T$ is isomorphic to $\mathbb{G}^d_m$ over $L$, then one says that $L$ splits $T$.

(3) We denote by $k^\ast$ the multiplicative group of non-zero elements of $k$ considered as an algebraic group over $k$. It is usually denoted by $G_m$ and is the affine algebraic group $Spec(k[t, t^{−1}])$ endowed with the comultiplication $t \to t \otimes t$ on the coordinate ring... An algebraic torus over $k$ is an algebraic group $T$ isomorphic to a finite direct product $k^\ast \times \dots \times k^\ast$.

(4) An algebraic torus is a type of commutative affine algebraic group. Let $F$ be a field with algebraic closure ${\overline {F}}$. Then a $F$-torus is an algebraic group defined over $F$ which is isomorphic over ${\overline {F}}$ to a finite product of copies of the multiplicative group. In other words, if $\mathbf {T}$ is an $F$-group it is a torus if and only if $\mathbf {T} ({\overline {F}})\cong ({\overline {F}}^{\times })^{r}$ for some $r\geq 1$.

(5) Let $G$ be a group. We say that $G$ is an algebraic group if $G$ is a quasi-projective variety and the two maps $m: G\times G \to G$ and $i: G \to G$, where $m$ is multiplication and $i$ is the inverse map, are both morphisms.

The group $\mathbb{G}_m$ is $GL_1(K)$. Note that as a group $\mathbb{G}_m$ is the set of units in $K$ under multiplication.

Let $G$ be an algebraic group. If $G$ is affine then we say that $G$ is a linear algebraic group. If $G$ is projective and connected then we say that $G$ is an abelian variety.

The algebaic group $\mathbb{G}^k_m$ is called a torus. So a torus in algebraic geometry is just a product of copies of $\mathbb{G}_m$.

  • $\begingroup$ FWIW, the definitions in (3) and (5) are, as stated here, of 'split' tori, so a priori not equivalent to the other defs: that is, 3 and 5 are special cases of others: I assume that in (3) and (5) the base fields are (separably/algebraically) closed? In any event, if the notation is confusing you: what do you know of alg. geo? 3 provides the std def of ${\mathbb G}_m$ over any field. A torus is an alg. group that, after base change (of fields, here), becomes iso to a (finite) product of ${\mathbb G}_m$. It turns out that one only needs a sep'ble ext, as in (1) and (2). $\endgroup$
    – peter a g
    May 15, 2018 at 12:54
  • $\begingroup$ Well I have heard that $SO(2)$ is a typical example of nonsplit torus defined over $\mathbb{R}$. Does anyone know how to check this by definition? $\endgroup$
    – Dick. Y
    Jan 31, 2021 at 8:06

1 Answer 1


Clasical definition: Let $H \subseteq \operatorname{GL}_n(\overline{k})$ be a connected linear algebraic group over $k$. So $H$ is a Zariski closed subgroup of $\operatorname{GL}_n(\overline{k})$ which is irreducible, and the polynomials of the radical ideal $I(H)$ corresponding to $H$ can be chosen to have coefficients in $k$, rather than just $\overline{k}$. We say that $H$ is an algebraic torus if there exists a $g \in \operatorname{GL}_n(\overline{k})$ such that $gHg^{-1}$ consists of diagonal matrices.

Group scheme definition: If $G$ is a group scheme over a field $k$, then the fibre product $G \times_k \operatorname{Spec}(\overline{k})$ is naturally a group scheme over $\overline{k}$. If $G_1$ and $G_2$ are group schemes over $k$, then so is the fibre product $G_1 \times_k G_2$ in a natural way.

Let $\mathbf G_m = \mathbf G_{m,k} = \operatorname{Spec}(k[t,t^{-1}])$. If $A$ is any $k$-algebra, then the set $$\mathbf G_m(A) := \operatorname{Hom}_{\textrm{k-sch}}(\operatorname{Spec} A, \mathbf G_m) = \operatorname{Hom}_{\textrm{k-alg}}(k[t,t^{-1}],A)$$ of $A$-rational points of $\mathbf G_m$ can be naturally identified with $A^{\ast}$, the group of units of $A$, because a $k$-algebra homomorphism $k[t,t^{-1}] \rightarrow A$ is completely determined by where it sends $t$, and it must send $t$ to a unit of $A$.

So $\mathbf G_m(A)$ has a natural group structure for each $k$-algebra $A$, and a homomorphism of $k$-algebras $\phi: A \rightarrow B$ induces a group homomorphism $\mathbf G_m(A) \rightarrow \mathbf G_m(B)$. By the Yoneda lemma, this uniquely makes $\mathbf G_m$ into a group scheme over $k$.

Putting this all together, since $\mathbf G_{m,\overline{k}} = \operatorname{Spec}(\overline{k}[t,t^{-1}])$ is a group scheme over $\overline{k}$, so is the fibre product $N = \prod\limits_{i=1}^n \mathbf G_{m,\overline{k}}$ over $\overline{k}$.

We say that a group scheme $G$ over $k$ is an algebraic torus if $G \times_k \operatorname{Spec}(\overline{k})$ is isomorphic to $N$ as $\overline{k}$-group schemes for some $n$.


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